Modular Forms and L-functions, Math 8207-8208
A course in modern number theory and harmonic analysis
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{displaystyle int _{0}^{1}int _{0}^{1}(xy)^{xy}\,dx\,dy=int _{0}^{1}x^{x}\,dx}{displaystyle int _{0}^{1}int _{0}^{1}(xy)^{xy}\,dx\,dy=int _{0}^{1}x^{x}\,dx}
Beweis
{displaystyle int _{0}^{1}int _{0}^{1}(xy)^{xy}\,dx\,dy}{displaystyle int _{0}^{1}int _{0}^{1}(xy)^{xy}\,dx\,dy} ist nach Substitution {displaystyle xmapsto {frac {x}{y}}}{displaystyle xmapsto {frac {x}{y}}} gleich {displaystyle int _{0}^{1}int _{0}^{y}x^{x}\,{frac {dx}{y}}\,dy}{displaystyle int _{0}^{1}int _{0}^{y}x^{x}\,{frac {dx}{y}}\,dy}.
Nach dem Vertauschen der Integrationsreihenfolge ist dies {displaystyle int _{0}^{1}int _{x}^{1}x^{x}\,{frac {dy}{y}}\,dx=int _{0}^{1}x^{x}\,(-log x)\,dx}{displaystyle int _{0}^{1}int _{x}^{1}x^{x}\,{frac {dy}{y}}\,dx=int _{0}^{1}x^{x}\,(-log x)\,dx}.
Wegen {displaystyle int _{0}^{1}x^{x}\,(1+log x)\,dx=left[x^{x} ight]_{0}^{1}=0}{displaystyle int _{0}^{1}x^{x}\,(1+log x)\,dx=left[x^{x} ight]_{0}^{1}=0} ist {displaystyle int _{0}^{1}x^{x}\,dx=int _{0}^{1}x^{x}\,(-log x)\,dx}{displaystyle int _{0}^{1}x^{x}\,dx=int _{0}^{1}x^{x}\,(-log x)\,dx}.
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{displaystyle int _{0}^{1}int _{0}^{1}{sqrt {x^{2}+y^{2}}}\,dx\,dy={frac {1}{3}}left({sqrt {2}}+log left({sqrt {2}}+1
ight)
ight)}{displaystyle int _{0}^{1}int _{0}^{1}{sqrt {x^{2}+y^{2}}}\,dx\,dy={frac {1}{3}}left({sqrt {2}}+log left({sqrt {2}}+1
ight)
ight)}
Beweis
Teile das Einheitsquadrat {displaystyle Q=[0,1]^{2}\,}{displaystyle Q=[0,1]^{2}\,} in zwei kongruente Dreiecke, jeweils mit dem Ursprung als Spitze.
{displaystyle Q_{1}={(x,y)in Q\,:\,xgeq y}\,,\,Q_{2}={(x,y)in Q\,:\,ygeq x}}{displaystyle Q_{1}={(x,y)in Q\,:\,xgeq y}\,,\,Q_{2}={(x,y)in Q\,:\,ygeq x}}
{displaystyle I:=iint _{Q}{sqrt {x^{2}+y^{2}}}\,dx\,dy=2iint _{Q_{1}}{sqrt {x^{2}+y^{2}}}\,dx\,dy}{displaystyle I:=iint _{Q}{sqrt {x^{2}+y^{2}}}\,dx\,dy=2iint _{Q_{1}}{sqrt {x^{2}+y^{2}}}\,dx\,dy}
Verwende nun Polarkoordinaten:
{displaystyle {egin{bmatrix}x=rcos varphi \y=rsin varphi \end{bmatrix}}quad qquad dx\,dy=r\,dr\,dvarphi quad qquad r={sqrt {x^{2}+y^{2}}}}{displaystyle {egin{bmatrix}x=rcos varphi \y=rsin varphi \end{bmatrix}}quad qquad dx\,dy=r\,dr\,dvarphi quad qquad r={sqrt {x^{2}+y^{2}}}}
Wegen
{displaystyle xgeq yiff rcos varphi geq rsin varphi iff cos varphi geq sin varphi iff 0leq varphi leq {frac {pi }{4}}}{displaystyle xgeq yiff rcos varphi geq rsin varphi iff cos varphi geq sin varphi iff 0leq varphi leq {frac {pi }{4}}}
{displaystyle xleq 1iff 0leq rleq {frac {1}{cos varphi }}}{displaystyle xleq 1iff 0leq rleq {frac {1}{cos varphi }}}
ist {displaystyle I=2int _{0}^{frac {pi }{4}}int _{0}^{frac {1}{cos varphi }}r^{2}\,dr\,dvarphi ={frac {2}{3}}int _{0}^{frac {pi }{4}}{frac {dvarphi }{cos ^{3}varphi }}}{displaystyle I=2int _{0}^{frac {pi }{4}}int _{0}^{frac {1}{cos varphi }}r^{2}\,dr\,dvarphi ={frac {2}{3}}int _{0}^{frac {pi }{4}}{frac {dvarphi }{cos ^{3}varphi }}}
{displaystyle ={frac {1}{3}}left[{frac {sin varphi }{cos ^{2}varphi }}+log(sec varphi + an varphi ) ight]_{0}^{frac {pi }{4}}={frac {1}{3}}left({sqrt {2}}+log left({sqrt {2}}+1 ight) ight)}{displaystyle ={frac {1}{3}}left[{frac {sin varphi }{cos ^{2}varphi }}+log(sec varphi + an varphi ) ight]_{0}^{frac {pi }{4}}={frac {1}{3}}left({sqrt {2}}+log left({sqrt {2}}+1 ight) ight)}.
3 Bearbeiten
{displaystyle int _{0}^{1}int _{0}^{1}int _{0}^{1}{sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=log({sqrt {3}}+1)-{frac {log 2}{2}}+{frac {sqrt {3}}{4}}-{frac {pi }{24}}}{displaystyle int _{0}^{1}int _{0}^{1}int _{0}^{1}{sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=log({sqrt {3}}+1)-{frac {log 2}{2}}+{frac {sqrt {3}}{4}}-{frac {pi }{24}}}
Beweis
Teile den Einheitswürfel {displaystyle V=[0,1]^{3}\,}{displaystyle V=[0,1]^{3}\,} in drei kongruente Pyramiden mit dem Ursprung als Spitze.
{displaystyle V_{1}={(x,y,z)in V\,:\,xgeq y,z}\,,\,V_{2}={(x,y,z)in V\,:\,ygeq x,z}\,,\,V_{3}={(x,y,z)in V\,:\,zgeq x,y}}{displaystyle V_{1}={(x,y,z)in V\,:\,xgeq y,z}\,,\,V_{2}={(x,y,z)in V\,:\,ygeq x,z}\,,\,V_{3}={(x,y,z)in V\,:\,zgeq x,y}}
{displaystyle I:=iiint _{V}{sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=3iiint _{V_{1}}{sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz}{displaystyle I:=iiint _{V}{sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=3iiint _{V_{1}}{sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz}
Verwende nun Kugelkoordinaten:
{displaystyle {egin{bmatrix}x=rsin heta cos varphi \y=rsin heta sin varphi \z=rcos heta end{bmatrix}}quad qquad dx\,dy\,dz=r^{2}sin heta \,dr\,d heta \,dvarphi quad qquad r={sqrt {x^{2}+y^{2}+z^{2}}}}{displaystyle {egin{bmatrix}x=rsin heta cos varphi \y=rsin heta sin varphi \z=rcos heta end{bmatrix}}quad qquad dx\,dy\,dz=r^{2}sin heta \,dr\,d heta \,dvarphi quad qquad r={sqrt {x^{2}+y^{2}+z^{2}}}}
Wegen
{displaystyle xgeq yiff rsin heta cos varphi geq rsin heta sin varphi iff cos varphi geq sin varphi iff 0leq varphi leq {frac {pi }{4}}}{displaystyle xgeq yiff rsin heta cos varphi geq rsin heta sin varphi iff cos varphi geq sin varphi iff 0leq varphi leq {frac {pi }{4}}}
{displaystyle xgeq ziff rsin heta cos varphi geq rcos heta iff an heta geq sec varphi iff arctan(sec varphi )leq heta leq {frac {pi }{2}}}{displaystyle xgeq ziff rsin heta cos varphi geq rcos heta iff an heta geq sec varphi iff arctan(sec varphi )leq heta leq {frac {pi }{2}}}
{displaystyle xleq 1iff 0leq rleq {frac {1}{sin heta \,cos varphi }}}{displaystyle xleq 1iff 0leq rleq {frac {1}{sin heta \,cos varphi }}}
ist {displaystyle I=3int _{0}^{frac {pi }{4}}int _{arctan(sec varphi )}^{frac {pi }{2}}int _{0}^{frac {1}{sin heta \,cos varphi }}rcdot r^{2}\,sin heta \,dr\,d heta \,dvarphi }{displaystyle I=3int _{0}^{frac {pi }{4}}int _{arctan(sec varphi )}^{frac {pi }{2}}int _{0}^{frac {1}{sin heta \,cos varphi }}rcdot r^{2}\,sin heta \,dr\,d heta \,dvarphi }
{displaystyle =3int _{0}^{frac {pi }{4}}int _{arctan(sec varphi )}^{frac {pi }{2}}left[{frac {r^{4}}{4}} ight]_{0}^{frac {1}{sin heta \,cos varphi }}\,sin heta \,d heta \,dvarphi ={frac {3}{4}}int _{0}^{frac {pi }{4}}int _{arctan(sec varphi )}^{frac {pi }{2}}{frac {d heta }{sin ^{3} heta }}\,{frac {dvarphi }{cos ^{4}varphi }}}{displaystyle =3int _{0}^{frac {pi }{4}}int _{arctan(sec varphi )}^{frac {pi }{2}}left[{frac {r^{4}}{4}} ight]_{0}^{frac {1}{sin heta \,cos varphi }}\,sin heta \,d heta \,dvarphi ={frac {3}{4}}int _{0}^{frac {pi }{4}}int _{arctan(sec varphi )}^{frac {pi }{2}}{frac {d heta }{sin ^{3} heta }}\,{frac {dvarphi }{cos ^{4}varphi }}}.
Substituiere {displaystyle heta ={ ext{arccot}}\,s}{displaystyle heta ={ ext{arccot}}\,s}, dann ist {displaystyle int _{arctan(sec varphi )}^{frac {pi }{2}}{frac {d heta }{sin ^{3} heta }}=int _{0}^{cos varphi }{sqrt {1+s^{2}}}\,ds=left[{frac {s}{2}}{sqrt {1+s^{2}}}+{frac {1}{2}}\,{ ext{arsinh}}\,s ight]_{0}^{cos varphi }}{displaystyle int _{arctan(sec varphi )}^{frac {pi }{2}}{frac {d heta }{sin ^{3} heta }}=int _{0}^{cos varphi }{sqrt {1+s^{2}}}\,ds=left[{frac {s}{2}}{sqrt {1+s^{2}}}+{frac {1}{2}}\,{ ext{arsinh}}\,s ight]_{0}^{cos varphi }}.
Also ist {displaystyle I={frac {3}{8}}int _{0}^{frac {pi }{4}}{frac {sqrt {1+cos ^{2}varphi }}{cos ^{3}varphi }}\,dvarphi +{frac {3}{8}}int _{0}^{frac {pi }{4}}{frac {{ ext{arsinh}}(cos varphi )}{cos ^{4}varphi }}\,dvarphi }{displaystyle I={frac {3}{8}}int _{0}^{frac {pi }{4}}{frac {sqrt {1+cos ^{2}varphi }}{cos ^{3}varphi }}\,dvarphi +{frac {3}{8}}int _{0}^{frac {pi }{4}}{frac {{ ext{arsinh}}(cos varphi )}{cos ^{4}varphi }}\,dvarphi },
wobei {displaystyle int _{0}^{frac {pi }{4}}{frac {sqrt {1+cos ^{2}varphi }}{cos ^{3}varphi }}\,dvarphi =int _{0}^{1}{frac {sqrt {1+{frac {1}{1+x^{2}}}}}{frac {1}{{sqrt {1+x^{2}}}^{3}}}}\,{frac {dx}{1+x^{2}}}=int _{0}^{1}{sqrt {2+x^{2}}}\,dx}{displaystyle int _{0}^{frac {pi }{4}}{frac {sqrt {1+cos ^{2}varphi }}{cos ^{3}varphi }}\,dvarphi =int _{0}^{1}{frac {sqrt {1+{frac {1}{1+x^{2}}}}}{frac {1}{{sqrt {1+x^{2}}}^{3}}}}\,{frac {dx}{1+x^{2}}}=int _{0}^{1}{sqrt {2+x^{2}}}\,dx}
{displaystyle =left[{frac {x}{2}}{sqrt {2+x^{2}}}+{ ext{arsinh}}{frac {x}{sqrt {2}}} ight]_{0}^{1}={frac {sqrt {3}}{2}}+{ ext{arsinh}}{frac {1}{sqrt {2}}}}{displaystyle =left[{frac {x}{2}}{sqrt {2+x^{2}}}+{ ext{arsinh}}{frac {x}{sqrt {2}}} ight]_{0}^{1}={frac {sqrt {3}}{2}}+{ ext{arsinh}}{frac {1}{sqrt {2}}}} ist. Und {displaystyle int _{0}^{frac {pi }{4}}{frac {{ ext{arsinh}}(cos varphi )}{cos ^{4}varphi }}\,dvarphi }{displaystyle int _{0}^{frac {pi }{4}}{frac {{ ext{arsinh}}(cos varphi )}{cos ^{4}varphi }}\,dvarphi }
{displaystyle =left[{ ext{arsinh}}(cos varphi )left({frac {1}{3}}\,{frac {sin varphi }{cos ^{3}varphi }}+{frac {2}{3}}{frac {sin varphi }{cos varphi }} ight) ight]_{0}^{frac {pi }{4}}+int _{0}^{frac {pi }{4}}{frac {sin varphi }{sqrt {1+cos ^{2}varphi }}}left({frac {1}{3}}\,{frac {sin varphi }{cos ^{3}varphi }}+{frac {2}{3}}{frac {sin varphi }{cos varphi }} ight)dvarphi }{displaystyle =left[{ ext{arsinh}}(cos varphi )left({frac {1}{3}}\,{frac {sin varphi }{cos ^{3}varphi }}+{frac {2}{3}}{frac {sin varphi }{cos varphi }} ight) ight]_{0}^{frac {pi }{4}}+int _{0}^{frac {pi }{4}}{frac {sin varphi }{sqrt {1+cos ^{2}varphi }}}left({frac {1}{3}}\,{frac {sin varphi }{cos ^{3}varphi }}+{frac {2}{3}}{frac {sin varphi }{cos varphi }} ight)dvarphi }
{displaystyle ={frac {4}{3}}\,{ ext{arsinh}}{frac {1}{sqrt {2}}}+{frac {1}{3}}int _{0}^{frac {pi }{4}}{frac {sin ^{2}varphi }{{sqrt {1+cos ^{2}varphi }}\,cos ^{3}varphi }}\,dvarphi +{frac {2}{3}}int _{0}^{frac {pi }{4}}{frac {sin ^{2}varphi }{{sqrt {1+cos ^{2}varphi }}\,cos varphi }}\,dvarphi }{displaystyle ={frac {4}{3}}\,{ ext{arsinh}}{frac {1}{sqrt {2}}}+{frac {1}{3}}int _{0}^{frac {pi }{4}}{frac {sin ^{2}varphi }{{sqrt {1+cos ^{2}varphi }}\,cos ^{3}varphi }}\,dvarphi +{frac {2}{3}}int _{0}^{frac {pi }{4}}{frac {sin ^{2}varphi }{{sqrt {1+cos ^{2}varphi }}\,cos varphi }}\,dvarphi },
wobei {displaystyle int _{0}^{frac {pi }{4}}{frac {sin ^{2}varphi }{{sqrt {1+cos ^{2}varphi }}\,cos ^{3}varphi }}\,dvarphi =int _{0}^{1}{frac {frac {x^{2}}{1+x^{2}}}{{sqrt {1+{frac {1}{1+x^{2}}}}}\,{frac {1}{{sqrt {1+x^{2}}}^{3}}}}}\,{frac {dx}{1+x^{2}}}=int _{0}^{1}{frac {x^{2}}{sqrt {2+x^{2}}}}\,dx}{displaystyle int _{0}^{frac {pi }{4}}{frac {sin ^{2}varphi }{{sqrt {1+cos ^{2}varphi }}\,cos ^{3}varphi }}\,dvarphi =int _{0}^{1}{frac {frac {x^{2}}{1+x^{2}}}{{sqrt {1+{frac {1}{1+x^{2}}}}}\,{frac {1}{{sqrt {1+x^{2}}}^{3}}}}}\,{frac {dx}{1+x^{2}}}=int _{0}^{1}{frac {x^{2}}{sqrt {2+x^{2}}}}\,dx}
{displaystyle =left[{frac {x}{2}}{sqrt {2+x^{2}}}-{ ext{arsinh}}{frac {x}{sqrt {2}}} ight]_{0}^{1}={frac {sqrt {3}}{2}}-{ ext{arsinh}}{frac {1}{sqrt {2}}}}{displaystyle =left[{frac {x}{2}}{sqrt {2+x^{2}}}-{ ext{arsinh}}{frac {x}{sqrt {2}}} ight]_{0}^{1}={frac {sqrt {3}}{2}}-{ ext{arsinh}}{frac {1}{sqrt {2}}}} ist,
und {displaystyle int _{0}^{frac {pi }{4}}{frac {sin ^{2}varphi }{{sqrt {1+cos ^{2}varphi }}\,cos varphi }}\,dvarphi =int _{0}^{1}{frac {frac {x^{2}}{1+x^{2}}}{{sqrt {1+{frac {1}{1+x^{2}}}}}\,{frac {1}{sqrt {1+x^{2}}}}}}\,{frac {dx}{1+x^{2}}}}{displaystyle int _{0}^{frac {pi }{4}}{frac {sin ^{2}varphi }{{sqrt {1+cos ^{2}varphi }}\,cos varphi }}\,dvarphi =int _{0}^{1}{frac {frac {x^{2}}{1+x^{2}}}{{sqrt {1+{frac {1}{1+x^{2}}}}}\,{frac {1}{sqrt {1+x^{2}}}}}}\,{frac {dx}{1+x^{2}}}}
{displaystyle =int _{0}^{1}{frac {x^{2}}{{sqrt {2+x^{2}}}\,(1+x^{2})}}\,dx=left[{ ext{arsinh}}{frac {x}{sqrt {2}}}-arctan {frac {x}{sqrt {2+x^{2}}}} ight]_{0}^{1}={ ext{arsinh}}{frac {1}{sqrt {2}}}-{frac {pi }{6}}}{displaystyle =int _{0}^{1}{frac {x^{2}}{{sqrt {2+x^{2}}}\,(1+x^{2})}}\,dx=left[{ ext{arsinh}}{frac {x}{sqrt {2}}}-arctan {frac {x}{sqrt {2+x^{2}}}} ight]_{0}^{1}={ ext{arsinh}}{frac {1}{sqrt {2}}}-{frac {pi }{6}}}.
Also ist {displaystyle I={frac {3}{8}}left({frac {sqrt {3}}{2}}+{ ext{arsinh}}{frac {1}{sqrt {2}}}
ight)+{frac {3}{8}}left({frac {4}{3}}\,{ ext{arsinh}}{frac {1}{sqrt {2}}}+{frac {1}{3}}left({frac {sqrt {3}}{2}}-{ ext{arsinh}}{frac {1}{sqrt {2}}}
ight)+{frac {2}{3}}left({ ext{arsinh}}{frac {1}{sqrt {2}}}-{frac {pi }{6}}
ight)
ight)}{displaystyle I={frac {3}{8}}left({frac {sqrt {3}}{2}}+{ ext{arsinh}}{frac {1}{sqrt {2}}}
ight)+{frac {3}{8}}left({frac {4}{3}}\,{ ext{arsinh}}{frac {1}{sqrt {2}}}+{frac {1}{3}}left({frac {sqrt {3}}{2}}-{ ext{arsinh}}{frac {1}{sqrt {2}}}
ight)+{frac {2}{3}}left({ ext{arsinh}}{frac {1}{sqrt {2}}}-{frac {pi }{6}}
ight)
ight)}
{displaystyle ={ ext{arsinh}}{frac {1}{sqrt {2}}}+{frac {sqrt {3}}{4}}-{frac {pi }{24}}}{displaystyle ={ ext{arsinh}}{frac {1}{sqrt {2}}}+{frac {sqrt {3}}{4}}-{frac {pi }{24}}}, wobei {displaystyle { ext{arsinh}}{frac {1}{sqrt {2}}}=log({sqrt {3}}+1)-{frac {log 2}{2}}}{displaystyle { ext{arsinh}}{frac {1}{sqrt {2}}}=log({sqrt {3}}+1)-{frac {log 2}{2}}} ist.
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{displaystyle int _{0}^{1}int _{0}^{1}max left{x^{alpha _{1}-1}\,y^{eta _{1}-1},x^{alpha _{2}-1}\,y^{eta _{2}-1}
ight}\,dx\,dy={frac {{frac {alpha _{1}-alpha _{2}}{alpha _{2}}}+{frac {eta _{2}-eta _{1}}{eta _{1}}}}{alpha _{1}eta _{2}-alpha _{2}eta _{1}}}qquad alpha _{1}>alpha _{2}>0\,,\,eta _{2}>eta _{1}>0}{displaystyle int _{0}^{1}int _{0}^{1}max left{x^{alpha _{1}-1}\,y^{eta _{1}-1},x^{alpha _{2}-1}\,y^{eta _{2}-1}
ight}\,dx\,dy={frac {{frac {alpha _{1}-alpha _{2}}{alpha _{2}}}+{frac {eta _{2}-eta _{1}}{eta _{1}}}}{alpha _{1}eta _{2}-alpha _{2}eta _{1}}}qquad alpha _{1}>alpha _{2}>0\,,\,eta _{2}>eta _{1}>0}
ohne Beweis
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Ist {displaystyle V=left{(x_{1},...,x_{n})geq 0\,left|\,left({frac {x_{1}}{a_{1}}}
ight)^{b_{1}}+...+left({frac {x_{n}}{a_{n}}}
ight)^{b_{n}}leq 1
ight.
ight}}{displaystyle V=left{(x_{1},...,x_{n})geq 0\,left|\,left({frac {x_{1}}{a_{1}}}
ight)^{b_{1}}+...+left({frac {x_{n}}{a_{n}}}
ight)^{b_{n}}leq 1
ight.
ight}}, so gilt
{displaystyle int _{V}x_{1}^{c_{1}-1}cdots x_{n}^{c_{n}-1}\,dx_{1}cdots dx_{n}={frac {{a_{1}}^{c_{1}}cdots {a_{n}}^{c_{n}}}{b_{1}cdots b_{n}}}\,{frac {Gamma left({frac {c_{1}}{b_{1}}}
ight)cdots Gamma left({frac {c_{n}}{b_{n}}}
ight)}{Gamma left({frac {c_{1}}{b_{1}}}+...+{frac {c_{n}}{b_{n}}}+1
ight)}}}{displaystyle int _{V}x_{1}^{c_{1}-1}cdots x_{n}^{c_{n}-1}\,dx_{1}cdots dx_{n}={frac {{a_{1}}^{c_{1}}cdots {a_{n}}^{c_{n}}}{b_{1}cdots b_{n}}}\,{frac {Gamma left({frac {c_{1}}{b_{1}}}
ight)cdots Gamma left({frac {c_{n}}{b_{n}}}
ight)}{Gamma left({frac {c_{1}}{b_{1}}}+...+{frac {c_{n}}{b_{n}}}+1
ight)}}}
ohne Beweis
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{displaystyle int _{0}^{1}cdots int _{0}^{1}max left{x_{1}^{alpha _{1}},...,x_{n}^{alpha _{n}}
ight}\,dx_{1}cdots dx_{n}={frac {{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}{1+{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}}qquad alpha _{1},...,alpha _{n}>0}{displaystyle int _{0}^{1}cdots int _{0}^{1}max left{x_{1}^{alpha _{1}},...,x_{n}^{alpha _{n}}
ight}\,dx_{1}cdots dx_{n}={frac {{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}{1+{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}}qquad alpha _{1},...,alpha _{n}>0}
Beweis
Es sei {displaystyle V=[0,1]^{n}\,}{displaystyle V=[0,1]^{n}\,} der {displaystyle n\,}n\,-dimensionale Einheitswürfel und {displaystyle V_{j}=left{(x_{1},...,x_{n})in V\,{Big |}\,x_{i}^{alpha _{i}}<x_{j}^{alpha _{j}}quad forall i
eq j
ight}}{displaystyle V_{j}=left{(x_{1},...,x_{n})in V\,{Big |}\,x_{i}^{alpha _{i}}<x_{j}^{alpha _{j}}quad forall i
eq j
ight}}.
Für alle Tupel {displaystyle (x_{1},...,x_{n})in V_{j}\,}{displaystyle (x_{1},...,x_{n})in V_{j}\,} ist dann {displaystyle x_{j}^{alpha _{j}}=max left{x_{1}^{alpha _{1}},...,x_{n}^{alpha _{n}} ight}}{displaystyle x_{j}^{alpha _{j}}=max left{x_{1}^{alpha _{1}},...,x_{n}^{alpha _{n}} ight}} und {displaystyle x_{i}<x_{j}^{alpha _{j}/alpha _{i}}\,\,forall i eq j}{displaystyle x_{i}<x_{j}^{alpha _{j}/alpha _{i}}\,\,forall i eq j}.
{displaystyle V\,}V\, ist die abgeschlossene Hülle der disjunkten Vereinigung aller {displaystyle V_{j}}{displaystyle V_{j}}'s. Daher ist {displaystyle int _{V}=int _{V_{1}}+...+int _{V_{n}}}{displaystyle int _{V}=int _{V_{1}}+...+int _{V_{n}}}.
{displaystyle int _{V_{j}}max left{x_{1}^{alpha _{1}},...,x_{n}^{alpha _{n}}
ight}dx_{1}cdots dx_{n}=int _{V_{j}}x_{j}^{alpha _{j}}dx_{1}cdots dx_{n}}{displaystyle int _{V_{j}}max left{x_{1}^{alpha _{1}},...,x_{n}^{alpha _{n}}
ight}dx_{1}cdots dx_{n}=int _{V_{j}}x_{j}^{alpha _{j}}dx_{1}cdots dx_{n}}
{displaystyle =int _{0}^{1}\,int _{0}^{x_{j}^{alpha _{j}/alpha _{n}}}cdots {widehat {int _{0}^{x_{j}^{alpha _{j}/alpha _{j}}}}}cdots int _{0}^{x_{j}^{alpha _{j}/alpha _{1}}}x_{j}^{alpha _{j}}\,dx_{1}cdots {widehat {dx_{j}}}cdots dx_{n}cdot dx_{j}}{displaystyle =int _{0}^{1}\,int _{0}^{x_{j}^{alpha _{j}/alpha _{n}}}cdots {widehat {int _{0}^{x_{j}^{alpha _{j}/alpha _{j}}}}}cdots int _{0}^{x_{j}^{alpha _{j}/alpha _{1}}}x_{j}^{alpha _{j}}\,dx_{1}cdots {widehat {dx_{j}}}cdots dx_{n}cdot dx_{j}}
{displaystyle =int _{0}^{1}x_{j}^{alpha _{j}}\,\,x_{j}^{alpha _{j}/alpha _{1}}cdots {widehat {x_{j}^{alpha _{j}/alpha _{j}}}}cdots x_{j}^{alpha _{j}/alpha _{n}}\,dx_{j}=int _{0}^{1}x_{j}^{alpha _{j}left(1+{frac {1}{alpha _{1}}}+...+{widehat {frac {1}{alpha _{j}}}}+...+{frac {1}{alpha _{n}}} ight)}\,dx_{j}}{displaystyle =int _{0}^{1}x_{j}^{alpha _{j}}\,\,x_{j}^{alpha _{j}/alpha _{1}}cdots {widehat {x_{j}^{alpha _{j}/alpha _{j}}}}cdots x_{j}^{alpha _{j}/alpha _{n}}\,dx_{j}=int _{0}^{1}x_{j}^{alpha _{j}left(1+{frac {1}{alpha _{1}}}+...+{widehat {frac {1}{alpha _{j}}}}+...+{frac {1}{alpha _{n}}} ight)}\,dx_{j}}
{displaystyle ={frac {1}{alpha _{j}left(1+{frac {1}{alpha _{1}}}+...+{widehat {frac {1}{alpha _{j}}}}+...+{frac {1}{alpha _{n}}} ight)+1}}={frac {frac {1}{alpha _{j}}}{1+{frac {1}{alpha _{1}}}+...+{widehat {frac {1}{alpha _{j}}}}+...+{frac {1}{alpha _{n}}}+{frac {1}{alpha _{j}}}}}={frac {frac {1}{alpha _{j}}}{1+{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}}}{displaystyle ={frac {1}{alpha _{j}left(1+{frac {1}{alpha _{1}}}+...+{widehat {frac {1}{alpha _{j}}}}+...+{frac {1}{alpha _{n}}} ight)+1}}={frac {frac {1}{alpha _{j}}}{1+{frac {1}{alpha _{1}}}+...+{widehat {frac {1}{alpha _{j}}}}+...+{frac {1}{alpha _{n}}}+{frac {1}{alpha _{j}}}}}={frac {frac {1}{alpha _{j}}}{1+{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}}}.
Also ist {displaystyle int _{0}^{1}cdots int _{0}^{1}max left{x_{1}^{alpha _{1}},...,x_{n}^{alpha _{n}} ight}\,dx_{1}cdots dx_{n}=int _{V_{1}}+...+int _{V_{n}}={frac {{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}{1+{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}}}{displaystyle int _{0}^{1}cdots int _{0}^{1}max left{x_{1}^{alpha _{1}},...,x_{n}^{alpha _{n}} ight}\,dx_{1}cdots dx_{n}=int _{V_{1}}+...+int _{V_{n}}={frac {{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}{1+{frac {1}{alpha _{1}}}+...+{frac {1}{alpha _{n}}}}}}.
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {(-log xy)^{s-2}}{1-xy}}\,(1-x)\,dx\,dy=Gamma (s)left(zeta (s)-{frac {1}{s-1}}
ight)qquad { ext{Re}}(s)>0}{displaystyle int _{0}^{1}int _{0}^{1}{frac {(-log xy)^{s-2}}{1-xy}}\,(1-x)\,dx\,dy=Gamma (s)left(zeta (s)-{frac {1}{s-1}}
ight)qquad { ext{Re}}(s)>0}
Beweis (Hadjicostas Formel)
{displaystyle I:=int _{0}^{1}int _{0}^{1}{frac {(-log xy)^{s-2}}{1-xy}}\,(1-x)\,dy\,dx}{displaystyle I:=int _{0}^{1}int _{0}^{1}{frac {(-log xy)^{s-2}}{1-xy}}\,(1-x)\,dy\,dx}
ist nach Substitution {displaystyle u=xy\,}{displaystyle u=xy\,} gleich {displaystyle int _{0}^{1}int _{0}^{x}{frac {(-log u)^{s-2}}{1-u}}\,(1-x)\,{frac {du}{x}}\,dx}{displaystyle int _{0}^{1}int _{0}^{x}{frac {(-log u)^{s-2}}{1-u}}\,(1-x)\,{frac {du}{x}}\,dx}.
Vertauscht man die Integrationsreihenfolge, so ist {displaystyle I=int _{0}^{1}int _{u}^{1}left({frac {1}{x}}-1 ight)dx\,{frac {(-log u)^{s-2}}{1-u}}\,du}{displaystyle I=int _{0}^{1}int _{u}^{1}left({frac {1}{x}}-1 ight)dx\,{frac {(-log u)^{s-2}}{1-u}}\,du}
{displaystyle =int _{0}^{1}left(-log u-(1-u) ight){frac {(-log u)^{s-2}}{1-u}}\,du=int _{0}^{1}left({frac {(-log u)^{s-1}}{1-u}}-(-log u)^{s-2} ight)du}{displaystyle =int _{0}^{1}left(-log u-(1-u) ight){frac {(-log u)^{s-2}}{1-u}}\,du=int _{0}^{1}left({frac {(-log u)^{s-1}}{1-u}}-(-log u)^{s-2} ight)du}.
Substituiert man {displaystyle u=e^{-t}\,}{displaystyle u=e^{-t}\,}, so ist {displaystyle I=int _{0}^{infty }left({frac {t^{s-1}}{e^{t}-1}}-t^{s-2}\,e^{-t} ight)dt}{displaystyle I=int _{0}^{infty }left({frac {t^{s-1}}{e^{t}-1}}-t^{s-2}\,e^{-t} ight)dt}.
Für {displaystyle { ext{Re}}(s)>1\,}{displaystyle { ext{Re}}(s)>1\,} ist das {displaystyle Gamma (s)zeta (s)-Gamma (s-1)=Gamma (s)left(zeta (s)-{frac {1}{s-1}} ight)}{displaystyle Gamma (s)zeta (s)-Gamma (s-1)=Gamma (s)left(zeta (s)-{frac {1}{s-1}} ight)}.
Wegen der analytischen Fortsetzbarkeit gilt dies auch für {displaystyle { ext{Re}}(s)>0\,}{displaystyle { ext{Re}}(s)>0\,},
wobei der Fall {displaystyle s=1\,}{displaystyle s=1\,} als Grenzwert {displaystyle gamma \,}{displaystyle gamma \,} zu interpretieren ist.
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{displaystyle int _{0}^{pi }int _{0}^{pi }int _{0}^{pi }{frac {dx\,dy\,dz}{1-cos x\,cos y\,cos z}}={frac {1}{4}}left[Gamma left({frac {1}{4}}
ight)
ight]^{4}}{displaystyle int _{0}^{pi }int _{0}^{pi }int _{0}^{pi }{frac {dx\,dy\,dz}{1-cos x\,cos y\,cos z}}={frac {1}{4}}left[Gamma left({frac {1}{4}}
ight)
ight]^{4}}
Beweis (Watson Integral)
{displaystyle I:=int _{0}^{pi }int _{0}^{pi }int _{0}^{pi }{frac {dx\,dy\,dz}{1-cos x\,cos y\,cos z}}}{displaystyle I:=int _{0}^{pi }int _{0}^{pi }int _{0}^{pi }{frac {dx\,dy\,dz}{1-cos x\,cos y\,cos z}}} ist nach den Substitutionen {displaystyle left[{egin{matrix}xmapsto 2arctan x\ymapsto 2arctan y\zmapsto 2arctan zend{matrix}}
ight]}{displaystyle left[{egin{matrix}xmapsto 2arctan x\ymapsto 2arctan y\zmapsto 2arctan zend{matrix}}
ight]} gleich
{displaystyle int _{0}^{infty }int _{0}^{infty }int _{0}^{infty }{frac {{frac {2dx}{1+x^{2}}}cdot {frac {2dy}{1+y^{2}}}cdot {frac {2dz}{1+z^{2}}}}{1-{frac {1-x^{2}}{1+x^{2}}}cdot {frac {1-y^{2}}{1+y^{2}}}cdot {frac {1-z^{2}}{1+z^{2}}}}}=8int _{0}^{infty }int _{0}^{infty }int _{0}^{infty }{frac {dx\,dy\,dz}{(1+x^{2})(1+y^{2})(1+z^{2})-(1-x^{2})(1-y^{2})(1-z^{2})}}}{displaystyle int _{0}^{infty }int _{0}^{infty }int _{0}^{infty }{frac {{frac {2dx}{1+x^{2}}}cdot {frac {2dy}{1+y^{2}}}cdot {frac {2dz}{1+z^{2}}}}{1-{frac {1-x^{2}}{1+x^{2}}}cdot {frac {1-y^{2}}{1+y^{2}}}cdot {frac {1-z^{2}}{1+z^{2}}}}}=8int _{0}^{infty }int _{0}^{infty }int _{0}^{infty }{frac {dx\,dy\,dz}{(1+x^{2})(1+y^{2})(1+z^{2})-(1-x^{2})(1-y^{2})(1-z^{2})}}}
{displaystyle =4int _{0}^{infty }int _{0}^{infty }int _{0}^{infty }{frac {dx\,dy\,dz}{x^{2}+y^{2}+z^{2}+x^{2}y^{2}z^{2}}}}{displaystyle =4int _{0}^{infty }int _{0}^{infty }int _{0}^{infty }{frac {dx\,dy\,dz}{x^{2}+y^{2}+z^{2}+x^{2}y^{2}z^{2}}}}.
Wechselt man von kartesischen Koordinaten zu Kugelkoordinaten {displaystyle left[{egin{matrix}x=r\,sin heta \,cos varphi \y=r\,sin heta \,sin varphi \z=r\,cos heta qquad ;end{matrix}} ight]}{displaystyle left[{egin{matrix}x=r\,sin heta \,cos varphi \y=r\,sin heta \,sin varphi \z=r\,cos heta qquad ;end{matrix}} ight]}, so ist
{displaystyle I=4int _{0}^{frac {pi }{2}}int _{0}^{frac {pi }{2}}int _{0}^{infty }{frac {r^{2}\,sin heta ;dr\,d heta \,dvarphi }{r^{2}+r^{2}\,sin ^{2} heta \,cos ^{2}varphi cdot r^{2}\,sin ^{2} heta \,sin ^{2}varphi cdot r^{2}\,cos ^{2} heta }}}{displaystyle I=4int _{0}^{frac {pi }{2}}int _{0}^{frac {pi }{2}}int _{0}^{infty }{frac {r^{2}\,sin heta ;dr\,d heta \,dvarphi }{r^{2}+r^{2}\,sin ^{2} heta \,cos ^{2}varphi cdot r^{2}\,sin ^{2} heta \,sin ^{2}varphi cdot r^{2}\,cos ^{2} heta }}}
{displaystyle =4int _{0}^{frac {pi }{2}}int _{0}^{frac {pi }{2}}int _{0}^{infty }{frac {dr}{1+{Big (}r\,sin heta \,{sqrt {cos heta }}\,{sqrt {cos varphi \,sin varphi }}{Big )}^{4}}}\,sin heta ;d heta \,dvarphi }{displaystyle =4int _{0}^{frac {pi }{2}}int _{0}^{frac {pi }{2}}int _{0}^{infty }{frac {dr}{1+{Big (}r\,sin heta \,{sqrt {cos heta }}\,{sqrt {cos varphi \,sin varphi }}{Big )}^{4}}}\,sin heta ;d heta \,dvarphi }
Das ist nach Substitution {displaystyle t=r\,sin heta \,{sqrt {cos heta }}\,{sqrt {sin varphi \,cos varphi }}}{displaystyle t=r\,sin heta \,{sqrt {cos heta }}\,{sqrt {sin varphi \,cos varphi }}} gleich {displaystyle 4int _{0}^{frac {pi }{2}}int _{0}^{frac {pi }{2}}int _{0}^{infty }{frac {dt\,d heta \,dvarphi }{(1+t^{4})\,{sqrt {cos heta }}\,{sqrt {sin varphi \,cos varphi }}}}}{displaystyle 4int _{0}^{frac {pi }{2}}int _{0}^{frac {pi }{2}}int _{0}^{infty }{frac {dt\,d heta \,dvarphi }{(1+t^{4})\,{sqrt {cos heta }}\,{sqrt {sin varphi \,cos varphi }}}}}
{displaystyle =4int _{0}^{infty }{frac {dt}{1+t^{4}}}cdot int _{0}^{frac {pi }{2}}{frac {d heta }{sqrt {cos heta }}}cdot int _{0}^{frac {pi }{2}}{frac {dvarphi }{sqrt {sin varphi \,cos varphi }}}=4cdot {frac {{sqrt {2}}\,pi }{4}}cdot {frac {Gamma ^{2}left({frac {1}{4}} ight)}{2\,{sqrt {2pi }}}}cdot {frac {Gamma ^{2}left({frac {1}{4}} ight)}{2{sqrt {pi }}}}={frac {1}{4}}\,left[Gamma left({frac {1}{4}} ight) ight]^{4}}{displaystyle =4int _{0}^{infty }{frac {dt}{1+t^{4}}}cdot int _{0}^{frac {pi }{2}}{frac {d heta }{sqrt {cos heta }}}cdot int _{0}^{frac {pi }{2}}{frac {dvarphi }{sqrt {sin varphi \,cos varphi }}}=4cdot {frac {{sqrt {2}}\,pi }{4}}cdot {frac {Gamma ^{2}left({frac {1}{4}} ight)}{2\,{sqrt {2pi }}}}cdot {frac {Gamma ^{2}left({frac {1}{4}} ight)}{2{sqrt {pi }}}}={frac {1}{4}}\,left[Gamma left({frac {1}{4}} ight) ight]^{4}}.
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{displaystyle int _{0}^{1}cdots int _{0}^{1}prod _{i=1}^{n}left(t_{i}^{alpha -1}\,(1-t_{i})^{eta -1}
ight)prod _{1leq i<jleq n}|t_{i}-t_{j}|^{2gamma }\,dt_{1}cdots dt_{n}=prod _{j=0}^{n-1}{frac {Gamma (alpha +jgamma )\,Gamma (eta +jgamma )\,Gamma (1+(j+1)gamma )}{Gamma (alpha +eta +(n+j-1)gamma )\,Gamma (1+gamma )}}}{displaystyle int _{0}^{1}cdots int _{0}^{1}prod _{i=1}^{n}left(t_{i}^{alpha -1}\,(1-t_{i})^{eta -1}
ight)prod _{1leq i<jleq n}|t_{i}-t_{j}|^{2gamma }\,dt_{1}cdots dt_{n}=prod _{j=0}^{n-1}{frac {Gamma (alpha +jgamma )\,Gamma (eta +jgamma )\,Gamma (1+(j+1)gamma )}{Gamma (alpha +eta +(n+j-1)gamma )\,Gamma (1+gamma )}}}
ohne Beweis (Selberg Integral)
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{displaystyle int _{0}^{infty }int _{0}^{infty }e^{-left(x+y+{frac {lambda ^{3}}{xy}}
ight)}cdot x^{{frac {1}{3}}-1}cdot y^{{frac {2}{3}}-1}\,dx\,dy={frac {2pi }{sqrt {3}}}\,e^{-3lambda }}{displaystyle int _{0}^{infty }int _{0}^{infty }e^{-left(x+y+{frac {lambda ^{3}}{xy}}
ight)}cdot x^{{frac {1}{3}}-1}cdot y^{{frac {2}{3}}-1}\,dx\,dy={frac {2pi }{sqrt {3}}}\,e^{-3lambda }}
Beweis (Formel von Liouville)
Setze {displaystyle R(lambda )=int _{0}^{infty }int _{0}^{infty }e^{-left(x+y+{frac {lambda ^{3}}{xy}}
ight)}cdot x^{{frac {1}{3}}-1}cdot y^{{frac {2}{3}}-1}\,dx\,dy}{displaystyle R(lambda )=int _{0}^{infty }int _{0}^{infty }e^{-left(x+y+{frac {lambda ^{3}}{xy}}
ight)}cdot x^{{frac {1}{3}}-1}cdot y^{{frac {2}{3}}-1}\,dx\,dy}.
Differenziere nach {displaystyle lambda \,}{displaystyle lambda \,} :
{displaystyle R'(lambda )=int _{0}^{infty }int _{0}^{infty }e^{-left(x+y+{frac {lambda ^{3}}{xy}} ight)}cdot {frac {3lambda ^{2}}{xy}}cdot x^{{frac {1}{3}}-1}cdot y^{{frac {2}{3}}-1}\,dx\,dy}{displaystyle R'(lambda )=int _{0}^{infty }int _{0}^{infty }e^{-left(x+y+{frac {lambda ^{3}}{xy}} ight)}cdot {frac {3lambda ^{2}}{xy}}cdot x^{{frac {1}{3}}-1}cdot y^{{frac {2}{3}}-1}\,dx\,dy}
Substituiere {displaystyle z={frac {lambda ^{3}}{xy}}\,Rightarrow \,x={frac {lambda ^{3}}{yz}}\,Rightarrow \,dx=-{frac {lambda ^{3}}{yz^{2}}}\,dz}{displaystyle z={frac {lambda ^{3}}{xy}}\,Rightarrow \,x={frac {lambda ^{3}}{yz}}\,Rightarrow \,dx=-{frac {lambda ^{3}}{yz^{2}}}\,dz} :
{displaystyle R'(lambda )=int _{0}^{infty }int _{infty }^{0}e^{-left({frac {lambda ^{3}}{yz}}+y+z ight)}cdot {frac {3z}{lambda }}cdot {frac {lambda ^{-2}}{y^{{frac {1}{3}}-1}cdot z^{{frac {1}{3}}-1}}}cdot y^{{frac {2}{3}}-1}cdot {frac {-lambda ^{3}}{yz^{2}}}\,dz\,dy}{displaystyle R'(lambda )=int _{0}^{infty }int _{infty }^{0}e^{-left({frac {lambda ^{3}}{yz}}+y+z ight)}cdot {frac {3z}{lambda }}cdot {frac {lambda ^{-2}}{y^{{frac {1}{3}}-1}cdot z^{{frac {1}{3}}-1}}}cdot y^{{frac {2}{3}}-1}cdot {frac {-lambda ^{3}}{yz^{2}}}\,dz\,dy}
{displaystyle =3int _{0}^{infty }int _{0}^{infty }e^{-left(y+z+{frac {lambda ^{3}}{yz}} ight)}cdot y^{{frac {1}{3}}-1}cdot z^{{frac {2}{3}}-1}\,dz\,dy=3cdot R(lambda )}{displaystyle =3int _{0}^{infty }int _{0}^{infty }e^{-left(y+z+{frac {lambda ^{3}}{yz}} ight)}cdot y^{{frac {1}{3}}-1}cdot z^{{frac {2}{3}}-1}\,dz\,dy=3cdot R(lambda )}
{displaystyle Rightarrow \,R(lambda )=R(0)cdot e^{-3lambda }}{displaystyle Rightarrow \,R(lambda )=R(0)cdot e^{-3lambda }} mit {displaystyle R(0)=int _{0}^{infty }int _{0}^{infty }e^{-(x+y)}cdot x^{{frac {1}{3}}-1}cdot y^{{frac {2}{3}}-1}\,dx\,dy}{displaystyle R(0)=int _{0}^{infty }int _{0}^{infty }e^{-(x+y)}cdot x^{{frac {1}{3}}-1}cdot y^{{frac {2}{3}}-1}\,dx\,dy}
{displaystyle =int _{0}^{infty }x^{{frac {1}{3}}-1}\,e^{-x}\,dxcdot int _{0}^{infty }y^{{frac {2}{3}}-1}\,e^{-y}\,dy=Gamma left({frac {1}{3}} ight)cdot Gamma left({frac {2}{3}} ight)={frac {pi }{sin {frac {pi }{3}}}}={frac {2pi }{sqrt {3}}}}{displaystyle =int _{0}^{infty }x^{{frac {1}{3}}-1}\,e^{-x}\,dxcdot int _{0}^{infty }y^{{frac {2}{3}}-1}\,e^{-y}\,dy=Gamma left({frac {1}{3}} ight)cdot Gamma left({frac {2}{3}} ight)={frac {pi }{sin {frac {pi }{3}}}}={frac {2pi }{sqrt {3}}}}
10.2 Bearbeiten
{displaystyle int _{0}^{infty }cdots int _{0}^{infty }e^{-left(x_{1}+x_{2}+...+x_{n-1}+{frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}}
ight)}\,x_{1}^{{frac {1}{n}}-1}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}cdots dx_{n-1}={frac {1}{sqrt {n}}}\,{sqrt {2pi }}^{\,n-1}\,e^{-nz}}int_0^infty cdots int_0^infty e^{-left(x_1+x_2+...+x_{n-1}+frac{z^n}{x_1\, x_2cdots x_{n-1}}
ight)}\, x_1^{frac{1}{n}-1}\, x_2^{frac{2}{n}-1} cdots x_{n-1}^{frac{n-1}{n}-1}\, dx_1\, dx_2 cdots dx_{n-1}=frac{1}{sqrt{n}}\, sqrt{2pi}^{\, n-1}\, e^{-nz}
Beweis (Formel von Liouville)
Setze {displaystyle F(z)=int _{0}^{infty }cdots int _{0}^{infty }e^{-left(x_{1}+x_{2}+...+x_{n-1}+{frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}}
ight)}\,x_{1}^{{frac {1}{n}}-1}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}cdots dx_{n-1}}{displaystyle F(z)=int _{0}^{infty }cdots int _{0}^{infty }e^{-left(x_{1}+x_{2}+...+x_{n-1}+{frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}}
ight)}\,x_{1}^{{frac {1}{n}}-1}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}cdots dx_{n-1}}.
{displaystyle F'(z)=int _{0}^{infty }cdots int _{0}^{infty }e^{-left(x_{1}+x_{2}+...+x_{n-1}+{frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}} ight)}\,{frac {-n\,z^{n-1}}{x_{1}\,x_{2}cdots x_{n-1}}}\,x_{1}^{{frac {1}{n}}-1}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}cdots dx_{n-1}}{displaystyle F'(z)=int _{0}^{infty }cdots int _{0}^{infty }e^{-left(x_{1}+x_{2}+...+x_{n-1}+{frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}} ight)}\,{frac {-n\,z^{n-1}}{x_{1}\,x_{2}cdots x_{n-1}}}\,x_{1}^{{frac {1}{n}}-1}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}cdots dx_{n-1}}
{displaystyle =n\,int _{0}^{infty }cdots int _{0}^{infty }e^{-left({frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}}+x_{2}+...+x_{n-1}+x_{1} ight)}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}cdot {frac {x_{1}^{frac {1}{n}}}{z}}cdot {frac {-z^{n}}{x_{1}^{2}\,x_{2}cdots x_{n-1}}}\,dx_{1}\,dx_{2}cdots dx_{n-1}}{displaystyle =n\,int _{0}^{infty }cdots int _{0}^{infty }e^{-left({frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}}+x_{2}+...+x_{n-1}+x_{1} ight)}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}cdot {frac {x_{1}^{frac {1}{n}}}{z}}cdot {frac {-z^{n}}{x_{1}^{2}\,x_{2}cdots x_{n-1}}}\,dx_{1}\,dx_{2}cdots dx_{n-1}}
Nach Substitution {displaystyle u_{1}={frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}}}{displaystyle u_{1}={frac {z^{n}}{x_{1}\,x_{2}cdots x_{n-1}}}} ist {displaystyle du_{1}={frac {-z^{n}}{x_{1}^{2}\,x_{2}cdots x_{n-1}}}\,dx_{1}}{displaystyle du_{1}={frac {-z^{n}}{x_{1}^{2}\,x_{2}cdots x_{n-1}}}\,dx_{1}} und wegen {displaystyle x_{1}={frac {z^{n}}{u_{1}cdot x_{2}cdots x_{n-1}}}}{displaystyle x_{1}={frac {z^{n}}{u_{1}cdot x_{2}cdots x_{n-1}}}} ist {displaystyle x_{1}^{frac {1}{n}}={frac {z}{u_{1}^{frac {1}{n}}\,x_{2}^{frac {1}{n}}cdots x_{n-1}^{frac {1}{n}}}}}{displaystyle x_{1}^{frac {1}{n}}={frac {z}{u_{1}^{frac {1}{n}}\,x_{2}^{frac {1}{n}}cdots x_{n-1}^{frac {1}{n}}}}}.
{displaystyle F'(z)=-n\,int _{0}^{infty }cdots int _{0}^{infty }e^{-left(u_{1}+x_{2}+...+x_{n-1}+{frac {z^{n}}{u_{1}\,x_{2}cdots x_{n-1}}} ight)}\,x_{2}^{{frac {1}{n}}-1}cdots x_{n-1}^{{frac {n-2}{n}}-1}\,u_{1}^{{frac {n-1}{n}}-1}\,du_{1}\,dx_{2}cdots dx_{n-1}}{displaystyle F'(z)=-n\,int _{0}^{infty }cdots int _{0}^{infty }e^{-left(u_{1}+x_{2}+...+x_{n-1}+{frac {z^{n}}{u_{1}\,x_{2}cdots x_{n-1}}} ight)}\,x_{2}^{{frac {1}{n}}-1}cdots x_{n-1}^{{frac {n-2}{n}}-1}\,u_{1}^{{frac {n-1}{n}}-1}\,du_{1}\,dx_{2}cdots dx_{n-1}}
Also ist {displaystyle F'(z)=-n\,F(z)}{displaystyle F'(z)=-n\,F(z)}, woraus {displaystyle F(z)=Ccdot e^{-nz}}{displaystyle F(z)=Ccdot e^{-nz}} folgt.
Und {displaystyle C=F(0)=int _{0}^{infty }cdots int _{0}^{infty }e^{-(x_{1}+x_{2}+...+x_{n-1})}\,x_{1}^{{frac {1}{n}}-1}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}cdots dx_{n-1}}{displaystyle C=F(0)=int _{0}^{infty }cdots int _{0}^{infty }e^{-(x_{1}+x_{2}+...+x_{n-1})}\,x_{1}^{{frac {1}{n}}-1}\,x_{2}^{{frac {2}{n}}-1}cdots x_{n-1}^{{frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}cdots dx_{n-1}}
{displaystyle =int _{0}^{infty }x_{1}^{{frac {1}{n}}-1}\,e^{-x_{1}}\,dx_{1}cdot int _{0}^{infty }x_{2}^{{frac {2}{n}}-1}\,e^{-x_{2}}\,dx_{2}cdots int _{0}^{infty }x_{n-1}^{{frac {n-1}{n}}-1}\,e^{-x_{n-1}}\,dx_{n-1}}{displaystyle =int _{0}^{infty }x_{1}^{{frac {1}{n}}-1}\,e^{-x_{1}}\,dx_{1}cdot int _{0}^{infty }x_{2}^{{frac {2}{n}}-1}\,e^{-x_{2}}\,dx_{2}cdots int _{0}^{infty }x_{n-1}^{{frac {n-1}{n}}-1}\,e^{-x_{n-1}}\,dx_{n-1}}
{displaystyle =Gamma left({frac {1}{n}} ight)cdot Gamma left({frac {2}{n}} ight)cdots Gamma left({frac {n-1}{n}} ight)={frac {1}{sqrt {n}}}\,{sqrt {2pi }}^{\,n-1}}{displaystyle =Gamma left({frac {1}{n}} ight)cdot Gamma left({frac {2}{n}} ight)cdots Gamma left({frac {n-1}{n}} ight)={frac {1}{sqrt {n}}}\,{sqrt {2pi }}^{\,n-1}}.
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{(1-xy)\,{sqrt {x\,(1-y)}}}}\,dx\,dy=8G}{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{(1-xy)\,{sqrt {x\,(1-y)}}}}\,dx\,dy=8G}
Beweis (Formel von Zudilin)
Substituiere {displaystyle u=xy}{displaystyle u=xy}:
{displaystyle I=int _{0}^{1}int _{0}^{y}{frac {1}{(1-u)\,{sqrt {{frac {u}{y}}\,(1-y)}}}}\,{frac {du}{y}}\,dy=int _{0}^{1}int _{0}^{y}{frac {du}{(1-u)\,{sqrt {u}}}}\,{frac {dy}{{sqrt {y}}\,{sqrt {1-y}}}}}{displaystyle I=int _{0}^{1}int _{0}^{y}{frac {1}{(1-u)\,{sqrt {{frac {u}{y}}\,(1-y)}}}}\,{frac {du}{y}}\,dy=int _{0}^{1}int _{0}^{y}{frac {du}{(1-u)\,{sqrt {u}}}}\,{frac {dy}{{sqrt {y}}\,{sqrt {1-y}}}}}
Vertausche die Integrationsreihenfolge:
{displaystyle I=int _{0}^{1}int _{u}^{1}{frac {dy}{{sqrt {y}}{sqrt {1-y}}}}\,{frac {du}{(1-u)\,{sqrt {u}}}}=int _{0}^{1}2arccos({sqrt {u}}\,)\,{frac {du}{(1-u)\,{sqrt {u}}}}}{displaystyle I=int _{0}^{1}int _{u}^{1}{frac {dy}{{sqrt {y}}{sqrt {1-y}}}}\,{frac {du}{(1-u)\,{sqrt {u}}}}=int _{0}^{1}2arccos({sqrt {u}}\,)\,{frac {du}{(1-u)\,{sqrt {u}}}}}
Substituiere {displaystyle u=w^{2}}{displaystyle u=w^{2}}:
{displaystyle I=int _{0}^{1}2arccos w\,{frac {2w\,dw}{(1-w^{2})\,w}}=4int _{0}^{1}{frac {arccos w}{1-w^{2}}}\,dw}{displaystyle I=int _{0}^{1}2arccos w\,{frac {2w\,dw}{(1-w^{2})\,w}}=4int _{0}^{1}{frac {arccos w}{1-w^{2}}}\,dw}
Substituiere {displaystyle w=cos t}{displaystyle w=cos t}:
{displaystyle I=4int _{frac {pi }{2}}^{0}{frac {t}{1-cos ^{2}t}}\,(-sin t)\,dt=4int _{0}^{frac {pi }{2}}{frac {t}{sin t}}\,dt}{displaystyle I=4int _{frac {pi }{2}}^{0}{frac {t}{1-cos ^{2}t}}\,(-sin t)\,dt=4int _{0}^{frac {pi }{2}}{frac {t}{sin t}}\,dt}
Substituiere {displaystyle t=2arctan s}{displaystyle t=2arctan s}:
{displaystyle I=8int _{0}^{1}{frac {arctan s}{frac {2s}{1+s^{2}}}}\,{frac {2ds}{1+s^{2}}}=8int _{0}^{1}{frac {arctan s}{s}}\,ds=8G}{displaystyle I=8int _{0}^{1}{frac {arctan s}{frac {2s}{1+s^{2}}}}\,{frac {2ds}{1+s^{2}}}=8int _{0}^{1}{frac {arctan s}{s}}\,ds=8G}
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{(x+y)\,{sqrt {(1-x)(1-y)}}}}\,dx\,dy=4G}{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{(x+y)\,{sqrt {(1-x)(1-y)}}}}\,dx\,dy=4G}
Beweis
{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{4\,(x+y)\,{sqrt {(1-x)(1-y)}}}}\,dx\,dy}{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{4\,(x+y)\,{sqrt {(1-x)(1-y)}}}}\,dx\,dy}
ist nach der Substitution {displaystyle x=1-u^{2}\,,\,y=1-v^{2}}{displaystyle x=1-u^{2}\,,\,y=1-v^{2}} gleich
{displaystyle int _{0}^{1}int _{0}^{1}{frac {2ucdot 2v}{4\,(2-u^{2}-v^{2})\,{sqrt {u^{2}v^{2}}}}}\,du\,dv=int _{0}^{1}int _{0}^{1}{frac {1}{2-u^{2}-v^{2}}}\,du\,dv=G}{displaystyle int _{0}^{1}int _{0}^{1}{frac {2ucdot 2v}{4\,(2-u^{2}-v^{2})\,{sqrt {u^{2}v^{2}}}}}\,du\,dv=int _{0}^{1}int _{0}^{1}{frac {1}{2-u^{2}-v^{2}}}\,du\,dv=G}.
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{sqrt {1+x^{2}+y^{2}}}}\,dx\,dy=log left(2+{sqrt {3}}
ight)-{frac {pi }{6}}}{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{sqrt {1+x^{2}+y^{2}}}}\,dx\,dy=log left(2+{sqrt {3}}
ight)-{frac {pi }{6}}}
ohne Beweis
14 Bearbeiten
{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=G}{displaystyle int _{0}^{1}int _{0}^{1}{frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=G}
Beweis
{displaystyle I:=int _{0}^{1}int _{0}^{1}{frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=2cdot int _{0}^{1}int _{0}^{y}{frac {1}{2-x^{2}-y^{2}}}\,dx\,dy}{displaystyle I:=int _{0}^{1}int _{0}^{1}{frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=2cdot int _{0}^{1}int _{0}^{y}{frac {1}{2-x^{2}-y^{2}}}\,dx\,dy}
Verwende Polarkoordinaten {displaystyle x=rcos varphi \,,\,y=rsin varphi }{displaystyle x=rcos varphi \,,\,y=rsin varphi }:
{displaystyle I=int _{0}^{frac {pi }{4}}int _{0}^{sec varphi }{frac {2r}{2-r^{2}}}\,dr\,dvarphi =int _{0}^{frac {pi }{4}}left[-log(2-r^{2}) ight]_{0}^{sec varphi }dvarphi =int _{0}^{frac {pi }{4}}left(log 2-log left(2-sec ^{2}varphi ight) ight)dvarphi }{displaystyle I=int _{0}^{frac {pi }{4}}int _{0}^{sec varphi }{frac {2r}{2-r^{2}}}\,dr\,dvarphi =int _{0}^{frac {pi }{4}}left[-log(2-r^{2}) ight]_{0}^{sec varphi }dvarphi =int _{0}^{frac {pi }{4}}left(log 2-log left(2-sec ^{2}varphi ight) ight)dvarphi }
{displaystyle ={frac {pi }{4}}log 2-int _{0}^{frac {pi }{4}}left[log left(2cos ^{2}varphi -1 ight)-log left(cos ^{2}varphi ight) ight]dvarphi }{displaystyle ={frac {pi }{4}}log 2-int _{0}^{frac {pi }{4}}left[log left(2cos ^{2}varphi -1 ight)-log left(cos ^{2}varphi ight) ight]dvarphi }, wobei {displaystyle 2cos ^{2}varphi -1=cos 2varphi }{displaystyle 2cos ^{2}varphi -1=cos 2varphi } ist.
Also ist {displaystyle I={frac {pi }{4}}log 2-underbrace {int _{0}^{frac {pi }{2}}log(cos varphi )\,{frac {dvarphi }{2}}} _{=-{frac {pi }{4}}log 2}+int _{0}^{frac {pi }{2}}log left(cos ^{2}{frac {varphi }{2}} ight){frac {dvarphi }{2}}={frac {pi }{2}}log 2+int _{0}^{frac {pi }{2}}log left(cos {frac {varphi }{2}} ight)dvarphi =G}{displaystyle I={frac {pi }{4}}log 2-underbrace {int _{0}^{frac {pi }{2}}log(cos varphi )\,{frac {dvarphi }{2}}} _{=-{frac {pi }{4}}log 2}+int _{0}^{frac {pi }{2}}log left(cos ^{2}{frac {varphi }{2}} ight){frac {dvarphi }{2}}={frac {pi }{2}}log 2+int _{0}^{frac {pi }{2}}log left(cos {frac {varphi }{2}} ight)dvarphi =G}.
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {1-x^{2}}{(1+x^{2}y^{2})\,log ^{2}(xy)}}\,dx\,dy={frac {2G}{pi }}-{frac {1}{2}}-log left(2\,\,{frac {Gamma left({frac {3}{4}}
ight)}{Gamma left({frac {1}{4}}
ight)}}
ight)}{displaystyle int _{0}^{1}int _{0}^{1}{frac {1-x^{2}}{(1+x^{2}y^{2})\,log ^{2}(xy)}}\,dx\,dy={frac {2G}{pi }}-{frac {1}{2}}-log left(2\,\,{frac {Gamma left({frac {3}{4}}
ight)}{Gamma left({frac {1}{4}}
ight)}}
ight)}
ohne Beweis
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{1+xy}}\,dx\,dy={frac {1}{2}}\,{frac {1}{alpha -eta }}left(left[psi left({frac {1}{2}}+{frac {eta }{2}}
ight)-psi left({frac {eta }{2}}
ight)
ight]-left[psi left({frac {1}{2}}+{frac {alpha }{2}}
ight)-psi left({frac {alpha }{2}}
ight)
ight]
ight)}{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{1+xy}}\,dx\,dy={frac {1}{2}}\,{frac {1}{alpha -eta }}left(left[psi left({frac {1}{2}}+{frac {eta }{2}}
ight)-psi left({frac {eta }{2}}
ight)
ight]-left[psi left({frac {1}{2}}+{frac {alpha }{2}}
ight)-psi left({frac {alpha }{2}}
ight)
ight]
ight)}
Beweis
{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{1+xy}}\,dx\,dy=sum _{k=0}^{infty }(-1)^{k}\,int _{0}^{1}int _{0}^{1}x^{k+alpha -1}\,y^{k+eta -1}\,dx\,dy=sum _{k=0}^{infty }(-1)^{k}\,{frac {1}{k+alpha }}cdot {frac {1}{k+eta }}}{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{1+xy}}\,dx\,dy=sum _{k=0}^{infty }(-1)^{k}\,int _{0}^{1}int _{0}^{1}x^{k+alpha -1}\,y^{k+eta -1}\,dx\,dy=sum _{k=0}^{infty }(-1)^{k}\,{frac {1}{k+alpha }}cdot {frac {1}{k+eta }}}
{displaystyle ={frac {1}{alpha -eta }}left(sum _{k=0}^{infty }{frac {(-1)^{k}}{k+eta }}-sum _{k=0}^{infty }{frac {(-1)^{k}}{k+alpha }} ight)}{displaystyle ={frac {1}{alpha -eta }}left(sum _{k=0}^{infty }{frac {(-1)^{k}}{k+eta }}-sum _{k=0}^{infty }{frac {(-1)^{k}}{k+alpha }} ight)} {displaystyle ={frac {1}{2}}\,{frac {1}{alpha -eta }}left(left[psi left({frac {1}{2}}+{frac {eta }{2}} ight)-psi left({frac {eta }{2}} ight) ight]-left[psi left({frac {1}{2}}+{frac {alpha }{2}} ight)-psi left({frac {alpha }{2}} ight) ight] ight)}{displaystyle ={frac {1}{2}}\,{frac {1}{alpha -eta }}left(left[psi left({frac {1}{2}}+{frac {eta }{2}} ight)-psi left({frac {eta }{2}} ight) ight]-left[psi left({frac {1}{2}}+{frac {alpha }{2}} ight)-psi left({frac {alpha }{2}} ight) ight] ight)}
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{(1+xy)\,log(xy)}}\,dx\,dy={frac {1}{alpha -eta }}log {frac {Gamma left({frac {alpha }{2}}
ight)\,Gamma left({frac {1}{2}}+{frac {eta }{2}}
ight)}{Gamma left({frac {eta }{2}}
ight)\,Gamma left({frac {1}{2}}+{frac {alpha }{2}}
ight)}}}{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{(1+xy)\,log(xy)}}\,dx\,dy={frac {1}{alpha -eta }}log {frac {Gamma left({frac {alpha }{2}}
ight)\,Gamma left({frac {1}{2}}+{frac {eta }{2}}
ight)}{Gamma left({frac {eta }{2}}
ight)\,Gamma left({frac {1}{2}}+{frac {alpha }{2}}
ight)}}}
Beweis
Aus der Formel {displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{1+xy}}\,dx\,dy={frac {1}{2}}\,{frac {1}{alpha -eta }}left(left[psi left({frac {1}{2}}+{frac {eta }{2}}
ight)-psi left({frac {eta }{2}}
ight)
ight]-left[psi left({frac {1}{2}}+{frac {alpha }{2}}
ight)-psi left({frac {alpha }{2}}
ight)
ight]
ight)}{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{1+xy}}\,dx\,dy={frac {1}{2}}\,{frac {1}{alpha -eta }}left(left[psi left({frac {1}{2}}+{frac {eta }{2}}
ight)-psi left({frac {eta }{2}}
ight)
ight]-left[psi left({frac {1}{2}}+{frac {alpha }{2}}
ight)-psi left({frac {alpha }{2}}
ight)
ight]
ight)} folgt
{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}\,(xy)^{gamma }}{1+xy}}\,dx\,dy={frac {1}{2}}\,{frac {1}{alpha -eta }}left(left[psi left({frac {1}{2}}+{frac {eta }{2}}+{frac {gamma }{2}} ight)-psi left({frac {eta }{2}}+{frac {gamma }{2}} ight) ight]-left[psi left({frac {1}{2}}+{frac {alpha }{2}}+{frac {gamma }{2}} ight)-psi left({frac {alpha }{2}}+{frac {gamma }{2}} ight) ight] ight)}{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}\,(xy)^{gamma }}{1+xy}}\,dx\,dy={frac {1}{2}}\,{frac {1}{alpha -eta }}left(left[psi left({frac {1}{2}}+{frac {eta }{2}}+{frac {gamma }{2}} ight)-psi left({frac {eta }{2}}+{frac {gamma }{2}} ight) ight]-left[psi left({frac {1}{2}}+{frac {alpha }{2}}+{frac {gamma }{2}} ight)-psi left({frac {alpha }{2}}+{frac {gamma }{2}} ight) ight] ight)}
Integriere nach {displaystyle gamma \,}{displaystyle gamma \,}:
{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}\,(xy)^{gamma }}{(1+xy)\,log(xy)}}\,dx\,dy={frac {1}{alpha -eta }}left(left[log Gamma left({frac {1}{2}}+{frac {eta }{2}}+{frac {gamma }{2}} ight)-log Gamma left({frac {eta }{2}}+{frac {gamma }{2}} ight) ight]-left[log Gamma left({frac {1}{2}}+{frac {alpha }{2}}+{frac {gamma }{2}} ight)-log Gamma left({frac {alpha }{2}}+{frac {gamma }{2}} ight) ight] ight)}{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}\,(xy)^{gamma }}{(1+xy)\,log(xy)}}\,dx\,dy={frac {1}{alpha -eta }}left(left[log Gamma left({frac {1}{2}}+{frac {eta }{2}}+{frac {gamma }{2}} ight)-log Gamma left({frac {eta }{2}}+{frac {gamma }{2}} ight) ight]-left[log Gamma left({frac {1}{2}}+{frac {alpha }{2}}+{frac {gamma }{2}} ight)-log Gamma left({frac {alpha }{2}}+{frac {gamma }{2}} ight) ight] ight)}
Setze {displaystyle gamma =0\,}{displaystyle gamma =0\,}:
{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{(1+xy)\,log(xy)}}\,dx\,dy={frac {1}{alpha -eta }}left(left[log Gamma left({frac {1}{2}}+{frac {eta }{2}} ight)-log Gamma left({frac {eta }{2}} ight) ight]-left[log Gamma left({frac {1}{2}}+{frac {alpha }{2}} ight)-log Gamma left({frac {alpha }{2}} ight) ight] ight)}{displaystyle int _{0}^{1}int _{0}^{1}{frac {x^{alpha -1}\,y^{eta -1}}{(1+xy)\,log(xy)}}\,dx\,dy={frac {1}{alpha -eta }}left(left[log Gamma left({frac {1}{2}}+{frac {eta }{2}} ight)-log Gamma left({frac {eta }{2}} ight) ight]-left[log Gamma left({frac {1}{2}}+{frac {alpha }{2}} ight)-log Gamma left({frac {alpha }{2}} ight) ight] ight)}
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {x}{(1+x^{2}y^{2})\,log(xy)}}\,dx\,dy=log left({sqrt {pi }}\,\,{frac {Gamma left({frac {3}{4}}
ight)}{Gamma left({frac {1}{4}}
ight)}}
ight)}{displaystyle int _{0}^{1}int _{0}^{1}{frac {x}{(1+x^{2}y^{2})\,log(xy)}}\,dx\,dy=log left({sqrt {pi }}\,\,{frac {Gamma left({frac {3}{4}}
ight)}{Gamma left({frac {1}{4}}
ight)}}
ight)}
ohne Beweis
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{displaystyle int _{0}^{1}int _{0}^{1}{frac {1-x^{2}}{(1+x^{2}y^{2})\,log ^{2}(xy)}}\,dx\,dy={frac {2G}{pi }}-{frac {1}{2}}-log left(2\,\,{frac {Gamma left({frac {3}{4}}
ight)}{Gamma left({frac {1}{4}}
ight)}}
ight)}{displaystyle int _{0}^{1}int _{0}^{1}{frac {1-x^{2}}{(1+x^{2}y^{2})\,log ^{2}(xy)}}\,dx\,dy={frac {2G}{pi }}-{frac {1}{2}}-log left(2\,\,{frac {Gamma left({frac {3}{4}}
ight)}{Gamma left({frac {1}{4}}
ight)}}
ight)}
Beweis
{displaystyle int _{0}^{1}int _{0}^{1}(1-x)cdot (xy)^{alpha -1}\,dx\,dy=int _{0}^{1}int _{0}^{1}x^{alpha -1}\,y^{alpha -1}\,dx\,dy-int _{0}^{1}int _{0}^{1}x^{alpha }\,y^{alpha -1}\,dx\,dy}{displaystyle int _{0}^{1}int _{0}^{1}(1-x)cdot (xy)^{alpha -1}\,dx\,dy=int _{0}^{1}int _{0}^{1}x^{alpha -1}\,y^{alpha -1}\,dx\,dy-int _{0}^{1}int _{0}^{1}x^{alpha }\,y^{alpha -1}\,dx\,dy}
{displaystyle ={frac {1}{alpha ^{2}}}-{frac {1}{alpha +1}}cdot {frac {1}{alpha }}={frac {1}{alpha ^{2}}}-{frac {1}{alpha }}+{frac {1}{alpha +1}}}{displaystyle ={frac {1}{alpha ^{2}}}-{frac {1}{alpha +1}}cdot {frac {1}{alpha }}={frac {1}{alpha ^{2}}}-{frac {1}{alpha }}+{frac {1}{alpha +1}}}
Integriere zweimal hintereinander nach {displaystyle alpha :}{displaystyle alpha :}
{displaystyle int _{0}^{1}int _{0}^{1}(1-x)\,{frac {(xy)^{alpha -1}}{log(xy)}}\,dx\,dy=-{frac {1}{alpha }}-log(alpha )+log(alpha +1)}{displaystyle int _{0}^{1}int _{0}^{1}(1-x)\,{frac {(xy)^{alpha -1}}{log(xy)}}\,dx\,dy=-{frac {1}{alpha }}-log(alpha )+log(alpha +1)}
{displaystyle int _{0}^{1}int _{0}^{1}(1-x)\,{frac {(xy)^{alpha -1}}{log ^{2}(xy)}}\,dx\,dy=(alpha +1)cdot log left(1+{frac {1}{alpha }} ight)-1}{displaystyle int _{0}^{1}int _{0}^{1}(1-x)\,{frac {(xy)^{alpha -1}}{log ^{2}(xy)}}\,dx\,dy=(alpha +1)cdot log left(1+{frac {1}{alpha }} ight)-1}
(Die Integrationskonstanten sind null, da beide Terme auf der rechten Seite gegen null gehen für {displaystyle alpha o infty }{displaystyle alpha o infty }.)
Substituiere {displaystyle xmapsto x^{2}\,,\,ymapsto y^{2}\,:}{displaystyle xmapsto x^{2}\,,\,ymapsto y^{2}\,:}
{displaystyle int _{0}^{1}int _{0}^{1}(1-x^{2})\,{frac {(xy)^{2alpha -1}}{log ^{2}(xy)}}\,dx\,dy=(alpha +1)cdot log left(1+{frac {1}{alpha }} ight)-1}{displaystyle int _{0}^{1}int _{0}^{1}(1-x^{2})\,{frac {(xy)^{2alpha -1}}{log ^{2}(xy)}}\,dx\,dy=(alpha +1)cdot log left(1+{frac {1}{alpha }} ight)-1}
Substituiere {displaystyle alpha ={frac {2k+1}{2}}\,:}{displaystyle alpha ={frac {2k+1}{2}}\,:}
{displaystyle int _{0}^{1}int _{0}^{1}(1-x^{2})\,{frac {(xy)^{2k}}{log ^{2}(xy)}}\,dx\,dy={frac {2k+3}{2}}cdot log left(1+{frac {2}{2k+1}} ight)-1}{displaystyle int _{0}^{1}int _{0}^{1}(1-x^{2})\,{frac {(xy)^{2k}}{log ^{2}(xy)}}\,dx\,dy={frac {2k+3}{2}}cdot log left(1+{frac {2}{2k+1}} ight)-1}
Generiere eine geometrische Reihe:
{displaystyle int _{0}^{1}int _{0}^{1}{frac {1-x^{2}}{(1+x^{2}y^{2})\,log ^{2}(xy)}}\,dx\,dy=sum _{k=0}^{infty }(-1)^{k}left[{frac {2k+3}{2}}cdot log left(1+{frac {2}{2k+1}} ight)-1 ight]}{displaystyle int _{0}^{1}int _{0}^{1}{frac {1-x^{2}}{(1+x^{2}y^{2})\,log ^{2}(xy)}}\,dx\,dy=sum _{k=0}^{infty }(-1)^{k}left[{frac {2k+3}{2}}cdot log left(1+{frac {2}{2k+1}} ight)-1 ight]}
{displaystyle =sum _{k=1}^{infty }(-1)^{k-1}left[{frac {2k+1}{2}}cdot log left({frac {2k+1}{2k-1}} ight)-1 ight]}{displaystyle =sum _{k=1}^{infty }(-1)^{k-1}left[{frac {2k+1}{2}}cdot log left({frac {2k+1}{2k-1}} ight)-1 ight]}
{displaystyle =underbrace {sum _{k=1}^{infty }(-1)^{k-1}left[2kcdot {frac {1}{2}}log left({frac {2k+1}{2k-1}} ight)-1 ight]} _{-{frac {1}{2}}+{frac {2G}{pi }}}+{frac {1}{2}}underbrace {sum _{k=1}^{infty }(-1)^{k-1}\,log left({frac {2k+1}{2k-1}} ight)} _{-2log left(2cdot {frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}}{displaystyle =underbrace {sum _{k=1}^{infty }(-1)^{k-1}left[2kcdot {frac {1}{2}}log left({frac {2k+1}{2k-1}} ight)-1 ight]} _{-{frac {1}{2}}+{frac {2G}{pi }}}+{frac {1}{2}}underbrace {sum _{k=1}^{infty }(-1)^{k-1}\,log left({frac {2k+1}{2k-1}} ight)} _{-2log left(2cdot {frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}}
Nun müssen noch die unterschweiften Ausdrücke nachgerechnet werden.
{displaystyle F(x):=sum _{k=1}^{infty }(-1)^{k-1}left[2kcdot {frac {1}{2}}log left({frac {1+{frac {x}{2k}}}{1-{frac {x}{2k}}}} ight)-x ight]}{displaystyle F(x):=sum _{k=1}^{infty }(-1)^{k-1}left[2kcdot {frac {1}{2}}log left({frac {1+{frac {x}{2k}}}{1-{frac {x}{2k}}}} ight)-x ight]}
{displaystyle =sum _{k=1}^{infty }(-1)^{k-1}sum _{n=1}^{infty }{frac {1}{2n+1}}\,{frac {x^{2n+1}}{(2k)^{2n}}}=sum _{n=1}^{infty }{frac {1}{2n+1}}sum _{k=1}^{infty }{frac {(-1)^{k-1}}{(2k)^{2n}}}\,x^{2n+1}=sum _{n=1}^{infty }{frac {1}{2n+1}}\,{frac {eta (2n)}{2^{2n}}}\,x^{2n+1}}{displaystyle =sum _{k=1}^{infty }(-1)^{k-1}sum _{n=1}^{infty }{frac {1}{2n+1}}\,{frac {x^{2n+1}}{(2k)^{2n}}}=sum _{n=1}^{infty }{frac {1}{2n+1}}sum _{k=1}^{infty }{frac {(-1)^{k-1}}{(2k)^{2n}}}\,x^{2n+1}=sum _{n=1}^{infty }{frac {1}{2n+1}}\,{frac {eta (2n)}{2^{2n}}}\,x^{2n+1}}
{displaystyle F'(x)=sum _{n=1}^{infty }eta (2n)\,left({frac {x}{2}} ight)^{2n}=-{frac {1}{2}}+{frac {1}{4}}cdot {frac {pi x}{sin {frac {pi x}{2}}}}}{displaystyle F'(x)=sum _{n=1}^{infty }eta (2n)\,left({frac {x}{2}} ight)^{2n}=-{frac {1}{2}}+{frac {1}{4}}cdot {frac {pi x}{sin {frac {pi x}{2}}}}}
Wegen {displaystyle F(0)=0}{displaystyle F(0)=0} ist {displaystyle F(1)=int _{0}^{1}F'(x)\,dx=-{frac {1}{2}}+{frac {1}{4}}int _{0}^{1}{frac {pi x}{sin {frac {pi x}{2}}}}\,dx=-{frac {1}{2}}+{frac {2G}{pi }}}{displaystyle F(1)=int _{0}^{1}F'(x)\,dx=-{frac {1}{2}}+{frac {1}{4}}int _{0}^{1}{frac {pi x}{sin {frac {pi x}{2}}}}\,dx=-{frac {1}{2}}+{frac {2G}{pi }}}.
{displaystyle sum _{k=1}^{infty }(-1)^{k-1}\,log left({frac {2k+1}{2k-1}} ight)=sum _{k=0}^{infty }(-1)^{k}\,log left({frac {2k+3}{2k+1}} ight)}{displaystyle sum _{k=1}^{infty }(-1)^{k-1}\,log left({frac {2k+1}{2k-1}} ight)=sum _{k=0}^{infty }(-1)^{k}\,log left({frac {2k+3}{2k+1}} ight)}
{displaystyle =sum _{k=0}^{infty }left(log {frac {4k+3}{4k+1}}-log {frac {4k+5}{4k+3}} ight)=sum _{k=0}^{infty }log {frac {(4k+3)(4k+3)}{(4k+1)(4k+5)}}}{displaystyle =sum _{k=0}^{infty }left(log {frac {4k+3}{4k+1}}-log {frac {4k+5}{4k+3}} ight)=sum _{k=0}^{infty }log {frac {(4k+3)(4k+3)}{(4k+1)(4k+5)}}}
{displaystyle =log prod _{k=0}^{infty }{frac {left(k+{frac {3}{4}} ight)left(k+{frac {3}{4}} ight)}{left(k+{frac {1}{4}} ight)left(k+{frac {5}{4}} ight)}}=log {frac {Gamma left({frac {1}{4}} ight)\,Gamma left({frac {5}{4}} ight)}{Gamma left({frac {3}{4}} ight)\,Gamma left({frac {3}{4}} ight)}}=log left({frac {1}{4}}cdot {frac {Gamma ^{2}left({frac {1}{4}} ight)}{Gamma ^{2}left({frac {3}{4}} ight)}} ight)}{displaystyle =log prod _{k=0}^{infty }{frac {left(k+{frac {3}{4}} ight)left(k+{frac {3}{4}} ight)}{left(k+{frac {1}{4}} ight)left(k+{frac {5}{4}} ight)}}=log {frac {Gamma left({frac {1}{4}} ight)\,Gamma left({frac {5}{4}} ight)}{Gamma left({frac {3}{4}} ight)\,Gamma left({frac {3}{4}} ight)}}=log left({frac {1}{4}}cdot {frac {Gamma ^{2}left({frac {1}{4}} ight)}{Gamma ^{2}left({frac {3}{4}} ight)}} ight)}
{displaystyle =2log left({frac {1}{2}}cdot {frac {Gamma left({frac {1}{4}} ight)}{Gamma left({frac {3}{4}} ight)}} ight)=-2log left(2cdot {frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}{displaystyle =2log left({frac {1}{2}}cdot {frac {Gamma left({frac {1}{4}} ight)}{Gamma left({frac {3}{4}} ight)}} ight)=-2log left(2cdot {frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}
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Ist {displaystyle V:=left{(x_{1},...,x_{n})geq 0\,{Big |}\,x_{1}+...+x_{n}=1
ight}}{displaystyle V:=left{(x_{1},...,x_{n})geq 0\,{Big |}\,x_{1}+...+x_{n}=1
ight}} und ist {displaystyle 0<a_{1}<...<a_{n}}{displaystyle 0<a_{1}<...<a_{n}}, so gilt
{displaystyle int limits _{V}{frac {1}{a_{1}x_{1}+...+a_{n}x_{n}}}\,dx_{1}cdots dx_{n}={frac {1}{(n-2)!}}sum _{k=1}^{n}{frac {log a_{k}}{a_{k}}}\,prod _{j=1 atop j
eq k}^{n}{frac {a_{k}}{a_{k}-a_{j}}}}{displaystyle int limits _{V}{frac {1}{a_{1}x_{1}+...+a_{n}x_{n}}}\,dx_{1}cdots dx_{n}={frac {1}{(n-2)!}}sum _{k=1}^{n}{frac {log a_{k}}{a_{k}}}\,prod _{j=1 atop j
eq k}^{n}{frac {a_{k}}{a_{k}-a_{j}}}}.
ohne Beweis
href{https://de.m.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Allgemeine_Integralformeln}{Formelsammlung Mathematik: Bestimmte Integrale: Allgemeine Integralformeln}
1.1 Bearbeiten
{displaystyle int _{-infty }^{infty }fleft(x-{frac {b}{x}}
ight)dx=int _{-infty }^{infty }f(x)\,dxqquad b>0}{displaystyle int _{-infty }^{infty }fleft(x-{frac {b}{x}}
ight)dx=int _{-infty }^{infty }f(x)\,dxqquad b>0}
Beweis (Formel nach Cauchy)
{displaystyle I:=int _{-infty }^{infty }fleft(x-{frac {b}{x}}
ight)dx=int _{-infty }^{0}fleft(x-{frac {b}{x}}
ight)dx+int _{0}^{infty }fleft(x-{frac {b}{x}}
ight)dx}{displaystyle I:=int _{-infty }^{infty }fleft(x-{frac {b}{x}}
ight)dx=int _{-infty }^{0}fleft(x-{frac {b}{x}}
ight)dx+int _{0}^{infty }fleft(x-{frac {b}{x}}
ight)dx}
Substituiert man bei beiden Integralen auf der rechten Seite {displaystyle xmapsto -{frac {b}{x}}}{displaystyle xmapsto -{frac {b}{x}}}, so ist
{displaystyle I=int _{0}^{infty }fleft(-{frac {b}{x}}+x ight){frac {b}{x^{2}}}\,dx+int _{-infty }^{0}fleft(-{frac {b}{x}}+x ight){frac {b}{x^{2}}}\,dx=int _{-infty }^{infty }fleft(x-{frac {b}{x}} ight){frac {b}{x^{2}}}\,dx}{displaystyle I=int _{0}^{infty }fleft(-{frac {b}{x}}+x ight){frac {b}{x^{2}}}\,dx+int _{-infty }^{0}fleft(-{frac {b}{x}}+x ight){frac {b}{x^{2}}}\,dx=int _{-infty }^{infty }fleft(x-{frac {b}{x}} ight){frac {b}{x^{2}}}\,dx}.
Daher ist
{displaystyle 2I=int _{-infty }^{infty }fleft(x-{frac {b}{x}} ight)left(1+{frac {b}{x^{2}}} ight)dx=int _{-infty }^{0}fleft(x-{frac {b}{x}} ight)left(1+{frac {b}{x^{2}}} ight)dx+int _{0}^{infty }fleft(x-{frac {b}{x}} ight)left(1+{frac {b}{x^{2}}} ight)dx}{displaystyle 2I=int _{-infty }^{infty }fleft(x-{frac {b}{x}} ight)left(1+{frac {b}{x^{2}}} ight)dx=int _{-infty }^{0}fleft(x-{frac {b}{x}} ight)left(1+{frac {b}{x^{2}}} ight)dx+int _{0}^{infty }fleft(x-{frac {b}{x}} ight)left(1+{frac {b}{x^{2}}} ight)dx}.
Substituiert man nun bei beiden Integralen auf der rechten Seite {displaystyle y=x-{frac {b}{x}};;left[dy=left(1+{frac {b}{x^{2}}} ight)dx ight]}{displaystyle y=x-{frac {b}{x}};;left[dy=left(1+{frac {b}{x^{2}}} ight)dx ight]}, so ist
{displaystyle 2I=int _{-infty }^{infty }f(y)dy+int _{-infty }^{infty }f(y)dy}{displaystyle 2I=int _{-infty }^{infty }f(y)dy+int _{-infty }^{infty }f(y)dy}, woraus die Behauptung folgt.
1.2 Bearbeiten
Ist {displaystyle phi (x):=x-sum _{k=0}^{n}{frac {a_{k}}{x+b_{k}}}}{displaystyle phi (x):=x-sum _{k=0}^{n}{frac {a_{k}}{x+b_{k}}}} mit {displaystyle a_{0},a_{1},...,a_{n}>0\,}{displaystyle a_{0},a_{1},...,a_{n}>0\,} und {displaystyle b_{0},b_{1},...,b_{n}in mathbb {R} }{displaystyle b_{0},b_{1},...,b_{n}in mathbb {R} }, so gilt {displaystyle int _{-infty }^{infty }f(phi (x))\,dx=int _{-infty }^{infty }f(x)\,dx}{displaystyle int _{-infty }^{infty }f(phi (x))\,dx=int _{-infty }^{infty }f(x)\,dx}.
ohne Beweis (Glassers Formel)
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Für ein {displaystyle 0<delta <1\,}{displaystyle 0<delta <1\,} sei {displaystyle H(delta ):=left{zin mathbb {C} \,|\,{ ext{Re}}(z)geq -delta
ight}}{displaystyle H(delta ):=left{zin mathbb {C} \,|\,{ ext{Re}}(z)geq -delta
ight}}.
{displaystyle f:\,H(delta ) o mathbb {C} }{displaystyle f:\,H(delta ) o mathbb {C} } sei eine analytische Funktion, zu der es Konstanten {displaystyle C,Pin mathbb {R} }{displaystyle C,Pin mathbb {R} } und {displaystyle A<pi \,}{displaystyle A<pi \,} gibt,
so dass {displaystyle |f(z)|leq Ccdot e^{Pcdot { ext{Re}}(z)+Acdot |{ ext{Im}}(z)|}}{displaystyle |f(z)|leq Ccdot e^{Pcdot { ext{Re}}(z)+Acdot |{ ext{Im}}(z)|}} gilt {displaystyle forall zin H(delta )}{displaystyle forall zin H(delta )}.
Wenn man für {displaystyle x>0\,}x>0\, und ein {displaystyle 0<c<delta qquad F(x):={frac {1}{2pi i}}int _{c-iinfty }^{c+iinfty }{frac {pi }{sin pi z}}\,f(-z)\,x^{-z}\,dz}{displaystyle 0<c<delta qquad F(x):={frac {1}{2pi i}}int _{c-iinfty }^{c+iinfty }{frac {pi }{sin pi z}}\,f(-z)\,x^{-z}\,dz} setzt, so ist
{displaystyle F(x)=sum _{k=0}^{infty }f(k)\,(-x)^{k}}{displaystyle F(x)=sum _{k=0}^{infty }f(k)\,(-x)^{k}} für {displaystyle 0<x<e^{-P}\,}{displaystyle 0<x<e^{-P}\,} und {displaystyle int _{0}^{infty }F(t)\,t^{s-1}\,dt={frac {pi }{sin pi s}}\,f(-s)}{displaystyle int _{0}^{infty }F(t)\,t^{s-1}\,dt={frac {pi }{sin pi s}}\,f(-s)} für {displaystyle 0<{ ext{Re}}(s)<delta \,}{displaystyle 0<{ ext{Re}}(s)<delta \,}.
ohne Beweis (Hardys Version von Ramanujan Master Theorem)
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Es sei {displaystyle f\,}f\, eine rationale Funktion ohne Polstellen auf der positiven reellen Achse und mit höchstens einer einfachen Polstelle im Ursprung.
Ist dann der Nennergrad um mindestens 2 größer als der Zählergrad, so gilt:
{displaystyle int _{0}^{infty }f(x)\,x^{alpha }\,dx={frac {pi }{sin alpha pi }}\,sum { ext{res}}{Big (}f(-z)\,z^{alpha }{Big )}qquad 0<alpha <1}{displaystyle int _{0}^{infty }f(x)\,x^{alpha }\,dx={frac {pi }{sin alpha pi }}\,sum { ext{res}}{Big (}f(-z)\,z^{alpha }{Big )}qquad 0<alpha <1}
ohne Beweis
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Ist {displaystyle f:mathbb {R} o mathbb {R} }{displaystyle f:mathbb {R} o mathbb {R} } integrierbar und {displaystyle pi \,}pi \,-periodisch, so gilt
{displaystyle int _{-infty }^{infty }f(x)\,{frac {sin x}{x}}\,dx=int _{0}^{pi }f(x)\,dx}{displaystyle int _{-infty }^{infty }f(x)\,{frac {sin x}{x}}\,dx=int _{0}^{pi }f(x)\,dx} und {displaystyle { ext{p.V.}}int _{-infty }^{infty }f(x)\,{frac { an x}{x}}\,dx=int _{0}^{pi }f(x)\,dx}{displaystyle { ext{p.V.}}int _{-infty }^{infty }f(x)\,{frac { an x}{x}}\,dx=int _{0}^{pi }f(x)\,dx}
Beweis (Formel von Lobatschewski)
{displaystyle int _{-infty }^{infty }f(x)\,{frac {sin x}{x}}\,dx=sum _{kin mathbb {Z} }int _{kpi }^{(k+1)pi }f(x)\,{frac {sin x}{x}}\,dx=sum _{kin mathbb {Z} }int _{0}^{pi }f(x)\,{frac {(-1)^{k}\,sin x}{x+kpi }}\,dx}{displaystyle int _{-infty }^{infty }f(x)\,{frac {sin x}{x}}\,dx=sum _{kin mathbb {Z} }int _{kpi }^{(k+1)pi }f(x)\,{frac {sin x}{x}}\,dx=sum _{kin mathbb {Z} }int _{0}^{pi }f(x)\,{frac {(-1)^{k}\,sin x}{x+kpi }}\,dx}
{displaystyle =int _{0}^{pi }f(x)underbrace {sum _{kin mathbb {Z} }{frac {(-1)^{k}}{x+kpi }}} _{csc x}\,sin x\,dx=int _{0}^{pi }f(x)\,dx}{displaystyle =int _{0}^{pi }f(x)underbrace {sum _{kin mathbb {Z} }{frac {(-1)^{k}}{x+kpi }}} _{csc x}\,sin x\,dx=int _{0}^{pi }f(x)\,dx}.
Analog ist
{displaystyle sum _{kin mathbb {Z} }int _{left(k-{frac {1}{2}} ight)pi +varepsilon }^{left(k+{frac {1}{2}} ight)pi -varepsilon }f(x)\,{frac { an x}{x}}\,dx=sum _{kin mathbb {Z} }int _{{frac {pi }{2}}+varepsilon }^{{frac {pi }{2}}-varepsilon }f(x)\,{frac { an x}{x+kpi }}\,dx=int _{{frac {pi }{2}}+varepsilon }^{{frac {pi }{2}}-varepsilon }f(x)\,underbrace {sum _{kin mathbb {Z} }{frac {1}{x+kpi }}} _{cot x}\, an x\,dx}{displaystyle sum _{kin mathbb {Z} }int _{left(k-{frac {1}{2}} ight)pi +varepsilon }^{left(k+{frac {1}{2}} ight)pi -varepsilon }f(x)\,{frac { an x}{x}}\,dx=sum _{kin mathbb {Z} }int _{{frac {pi }{2}}+varepsilon }^{{frac {pi }{2}}-varepsilon }f(x)\,{frac { an x}{x+kpi }}\,dx=int _{{frac {pi }{2}}+varepsilon }^{{frac {pi }{2}}-varepsilon }f(x)\,underbrace {sum _{kin mathbb {Z} }{frac {1}{x+kpi }}} _{cot x}\, an x\,dx}
{displaystyle =int _{{frac {pi }{2}}+varepsilon }^{{frac {pi }{2}}-varepsilon }f(x)\,dx o int _{-{frac {pi }{2}}}^{frac {pi }{2}}f(x)\,dx=int _{0}^{pi }f(x)\,dx}{displaystyle =int _{{frac {pi }{2}}+varepsilon }^{{frac {pi }{2}}-varepsilon }f(x)\,dx o int _{-{frac {pi }{2}}}^{frac {pi }{2}}f(x)\,dx=int _{0}^{pi }f(x)\,dx} für {displaystyle varepsilon o 0+}varepsilon o 0+.
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Für {displaystyle |R|>|r|geq 0}{displaystyle |R|>|r|geq 0} und {displaystyle phi in mathbb {C} }{displaystyle phi in mathbb {C} }, mit {displaystyle left|{ ext{Im}}phi
ight|<log left|{frac {R}{r}}
ight|}{displaystyle left|{ ext{Im}}phi
ight|<log left|{frac {R}{r}}
ight|} falls {displaystyle r
eq 0\,}{displaystyle r
eq 0\,} ist, sei der Poissonsche Integralkern {displaystyle P_{R}(r,phi )\,}{displaystyle P_{R}(r,phi )\,} definiert als {displaystyle {frac {R^{2}-r^{2}}{R^{2}-2Rrcos phi +r^{2}}}}{displaystyle {frac {R^{2}-r^{2}}{R^{2}-2Rrcos phi +r^{2}}}}.
Ist {displaystyle f:{overline {B_{R}(0)}} o mathbb {C} ;,;zmapsto sum _{k=0}^{infty }a_{k}z^{k}}{displaystyle f:{overline {B_{R}(0)}} o mathbb {C} ;,;zmapsto sum _{k=0}^{infty }a_{k}z^{k}} eine holomorphe Funktion, so gilt {displaystyle {frac {1}{2pi }}int _{-pi }^{pi }P_{R}(r,phi -varphi )\,f(Re^{ivarphi })\,dvarphi =f(re^{iphi })}{displaystyle {frac {1}{2pi }}int _{-pi }^{pi }P_{R}(r,phi -varphi )\,f(Re^{ivarphi })\,dvarphi =f(re^{iphi })}.
Beweis (Poissonsche Integralformel)
Der Kern {displaystyle P_{R}(r,phi )\,}{displaystyle P_{R}(r,phi )\,} besitzt die komplexe Fourierreihenentwicklung {displaystyle sum _{nin mathbb {Z} }left({frac {r}{R}}
ight)^{|n|}e^{inphi }}{displaystyle sum _{nin mathbb {Z} }left({frac {r}{R}}
ight)^{|n|}e^{inphi }}.
Nun ist {displaystyle int _{-pi }^{pi }P_{R}(r,phi -varphi )\,f(Re^{ivarphi })\,dvarphi =int _{-pi }^{pi }sum _{nin mathbb {Z} }left({frac {r}{R}} ight)^{|n|}e^{in(phi -varphi )}sum _{k=0}^{infty }a_{k}R^{k}e^{ikvarphi }\,;dvarphi }{displaystyle int _{-pi }^{pi }P_{R}(r,phi -varphi )\,f(Re^{ivarphi })\,dvarphi =int _{-pi }^{pi }sum _{nin mathbb {Z} }left({frac {r}{R}} ight)^{|n|}e^{in(phi -varphi )}sum _{k=0}^{infty }a_{k}R^{k}e^{ikvarphi }\,;dvarphi }
{displaystyle =sum _{k=0}^{infty }sum _{nin mathbb {Z} }a_{k}R^{k}left({frac {r}{R}} ight)^{|n|}e^{inphi }underbrace {int _{-pi }^{pi }e^{-invarphi }\,e^{ikvarphi }\,dvarphi } _{2pi delta _{nk}}=2pi sum _{k=0}^{infty }a_{k}r^{k}e^{ikphi }=2pi \,f(re^{iphi })}{displaystyle =sum _{k=0}^{infty }sum _{nin mathbb {Z} }a_{k}R^{k}left({frac {r}{R}} ight)^{|n|}e^{inphi }underbrace {int _{-pi }^{pi }e^{-invarphi }\,e^{ikvarphi }\,dvarphi } _{2pi delta _{nk}}=2pi sum _{k=0}^{infty }a_{k}r^{k}e^{ikphi }=2pi \,f(re^{iphi })}.
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{displaystyle int _{0}^{infty }fleft(x+{frac {alpha }{x}}
ight)dx=int _{0}^{infty }fleft({sqrt {x^{2}+4alpha }}\,
ight)dxqquad alpha >0}{displaystyle int _{0}^{infty }fleft(x+{frac {alpha }{x}}
ight)dx=int _{0}^{infty }fleft({sqrt {x^{2}+4alpha }}\,
ight)dxqquad alpha >0}
Beweis
{displaystyle I:=int _{0}^{infty }fleft(x+{frac {alpha }{x}}
ight)dx=underbrace {int limits _{0}^{sqrt {alpha }}fleft(u+{frac {alpha }{u}}
ight)du} _{=:J}+int limits _{sqrt {a}}^{infty }fleft(t+{frac {alpha }{t}}
ight)dt}{displaystyle I:=int _{0}^{infty }fleft(x+{frac {alpha }{x}}
ight)dx=underbrace {int limits _{0}^{sqrt {alpha }}fleft(u+{frac {alpha }{u}}
ight)du} _{=:J}+int limits _{sqrt {a}}^{infty }fleft(t+{frac {alpha }{t}}
ight)dt}
Nach Substitution {displaystyle u={frac {alpha }{t}}}{displaystyle u={frac {alpha }{t}}} ist {displaystyle J=int limits _{infty }^{sqrt {a}}fleft({frac {alpha }{t}}+t ight){frac {-alpha }{t^{2}}}\,dt=int limits _{sqrt {a}}^{infty }fleft(t+{frac {alpha }{t}} ight)\,{frac {alpha }{t^{2}}}\,dt}{displaystyle J=int limits _{infty }^{sqrt {a}}fleft({frac {alpha }{t}}+t ight){frac {-alpha }{t^{2}}}\,dt=int limits _{sqrt {a}}^{infty }fleft(t+{frac {alpha }{t}} ight)\,{frac {alpha }{t^{2}}}\,dt}.
Also ist {displaystyle I=int limits _{sqrt {a}}^{infty }fleft(t+{frac {alpha }{t}} ight)left(1+{frac {alpha }{t^{2}}} ight)dt=int limits _{sqrt {a}}^{infty }fleft({sqrt {left(t-{frac {alpha }{t}} ight)^{2}+4alpha }}\, ight)cdot left(1+{frac {alpha }{t^{2}}} ight)dt}{displaystyle I=int limits _{sqrt {a}}^{infty }fleft(t+{frac {alpha }{t}} ight)left(1+{frac {alpha }{t^{2}}} ight)dt=int limits _{sqrt {a}}^{infty }fleft({sqrt {left(t-{frac {alpha }{t}} ight)^{2}+4alpha }}\, ight)cdot left(1+{frac {alpha }{t^{2}}} ight)dt}.
Nach Substitution {displaystyle x=t-{frac {alpha }{t}}}{displaystyle x=t-{frac {alpha }{t}}} ist {displaystyle I=int _{0}^{infty }fleft({sqrt {x^{2}+4alpha }}\, ight)dx}{displaystyle I=int _{0}^{infty }fleft({sqrt {x^{2}+4alpha }}\, ight)dx}.
href{https://de.m.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R(x,exp)}{Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,exp)}
0.1 Bearbeiten
{displaystyle int _{-infty }^{infty }e^{-x^{2}}\,dx={sqrt {pi }}}{displaystyle int _{-infty }^{infty }e^{-x^{2}}\,dx={sqrt {pi }}}
1. Beweis
{displaystyle I^{2}=left(int _{-infty }^{infty }e^{-x^{2}}\,dx
ight)\,left(int _{-infty }^{infty }e^{-y^{2}}\,dy
ight)=int _{-infty }^{infty }int _{-infty }^{infty }e^{-(x^{2}+y^{2})}\,dx\,dy}{displaystyle I^{2}=left(int _{-infty }^{infty }e^{-x^{2}}\,dx
ight)\,left(int _{-infty }^{infty }e^{-y^{2}}\,dy
ight)=int _{-infty }^{infty }int _{-infty }^{infty }e^{-(x^{2}+y^{2})}\,dx\,dy}
lässt sich in Polarkoordinaten schreiben als {displaystyle int _{0}^{2pi }int _{0}^{infty }e^{-r^{2}}\,r\,dr\,dvarphi }{displaystyle int _{0}^{2pi }int _{0}^{infty }e^{-r^{2}}\,r\,dr\,dvarphi }.
Und das ist {displaystyle int _{0}^{2pi }left[-{frac {e^{-r^{2}}}{2}} ight]_{0}^{infty }\,dvarphi =int _{0}^{2pi }{frac {1}{2}}\,dvarphi =pi Rightarrow I={sqrt {pi }}}{displaystyle int _{0}^{2pi }left[-{frac {e^{-r^{2}}}{2}} ight]_{0}^{infty }\,dvarphi =int _{0}^{2pi }{frac {1}{2}}\,dvarphi =pi Rightarrow I={sqrt {pi }}}.
2. Beweis
Die Fläche, die entsteht wenn {displaystyle f(x)=e^{-x^{2}}}{displaystyle f(x)=e^{-x^{2}}} um die z-Achse rotiert, schließt mit der xy-Ebene das gleiche Volumen ein
wie die Fläche, die entsteht, wenn {displaystyle f^{-1}(x)={sqrt {-log x}}}{displaystyle f^{-1}(x)={sqrt {-log x}}} um die x-Achse rotiert, mit der yz-Ebene.
Also {displaystyle I^{2}=pi int _{0}^{1}{sqrt {-log x}}^{\,2}\,dx=pi Rightarrow I={sqrt {pi }}}{displaystyle I^{2}=pi int _{0}^{1}{sqrt {-log x}}^{\,2}\,dx=pi Rightarrow I={sqrt {pi }}}.
3. Beweis
Definiert man {displaystyle F(x)=left(int _{0}^{x}e^{-t^{2}}\,dt
ight)^{2}}{displaystyle F(x)=left(int _{0}^{x}e^{-t^{2}}\,dt
ight)^{2}} und {displaystyle G(x)=int _{0}^{1}{frac {e^{-x^{2}\,(1+t^{2})}}{1+t^{2}}}\,dt}{displaystyle G(x)=int _{0}^{1}{frac {e^{-x^{2}\,(1+t^{2})}}{1+t^{2}}}\,dt}, so ist {displaystyle F'(x)=2\,int _{0}^{x}e^{-t^{2}}\,dt;e^{-x^{2}}}{displaystyle F'(x)=2\,int _{0}^{x}e^{-t^{2}}\,dt;e^{-x^{2}}}
und {displaystyle G'(x)=int _{0}^{1}e^{-x^{2}\,(1+t^{2})}\,(-2x)\,dt=-2\,int _{0}^{1}e^{-x^{2}\,t^{2}}\,x\,dt;e^{-x^{2}}=-2int _{0}^{x}e^{-t^{2}}\,dt;e^{-x^{2}}}{displaystyle G'(x)=int _{0}^{1}e^{-x^{2}\,(1+t^{2})}\,(-2x)\,dt=-2\,int _{0}^{1}e^{-x^{2}\,t^{2}}\,x\,dt;e^{-x^{2}}=-2int _{0}^{x}e^{-t^{2}}\,dt;e^{-x^{2}}}.
Es ist also {displaystyle F'(x)+G'(x)=0\,}{displaystyle F'(x)+G'(x)=0\,}. Folglich muss {displaystyle F(x)+G(x)\,}{displaystyle F(x)+G(x)\,} konstant sein.
{displaystyle F(infty )+G(infty )=F(0)+G(0)Rightarrow left({frac {I}{2}} ight)^{2}+0=0+{frac {pi }{4}}Rightarrow I={sqrt {pi }}}{displaystyle F(infty )+G(infty )=F(0)+G(0)Rightarrow left({frac {I}{2}} ight)^{2}+0=0+{frac {pi }{4}}Rightarrow I={sqrt {pi }}}
4. Beweis
Es sei {displaystyle a={sqrt {ipi }}=(1+i){sqrt {frac {pi }{2}}}}{displaystyle a={sqrt {ipi }}=(1+i){sqrt {frac {pi }{2}}}} und {displaystyle f(z)={frac {e^{-z^{2}}}{1+e^{-2az}}}}{displaystyle f(z)={frac {e^{-z^{2}}}{1+e^{-2az}}}}. Gaussintegralberechnung.PNG
Wegen {displaystyle { ext{Re}}(a)>0\,}{displaystyle { ext{Re}}(a)>0\,} gilt {displaystyle e^{-2ax} o left{{egin{matrix}0&,&x o infty \infty &,&x o -infty end{matrix}} ight.}{displaystyle e^{-2ax} o left{{egin{matrix}0&,&x o infty \infty &,&x o -infty end{matrix}} ight.}.
Ist {displaystyle 0leq yleq {sqrt {frac {pi }{2}}}}{displaystyle 0leq yleq {sqrt {frac {pi }{2}}}}, so geht für {displaystyle x o pm infty \,}{displaystyle x o pm infty \,} der Nenner von {displaystyle f(x+iy)={frac {e^{-x^{2}}\,e^{y^{2}-2ixy}}{1+e^{-2ax}\,e^{-2iay}}}}{displaystyle f(x+iy)={frac {e^{-x^{2}}\,e^{y^{2}-2ixy}}{1+e^{-2ax}\,e^{-2iay}}}} gegen {displaystyle 1\,}1\, oder {displaystyle infty \,}{displaystyle infty \,}
und der Zähler geht gegen Null. Also verschwinden die beiden Integrale {displaystyle int _{pm R}^{pm R+a}\,f\,dz}{displaystyle int _{pm R}^{pm R+a}\,f\,dz} für {displaystyle R o infty \,}R o infty \,.
Wegen {displaystyle f(z)-f(z+a)=e^{-z^{2}}}{displaystyle f(z)-f(z+a)=e^{-z^{2}}} gilt nun {displaystyle int _{-infty }^{infty }e^{-z^{2}}\,dz=lim _{R o infty }oint f\,dz=2pi i\,{ ext{res}}left(f,{frac {a}{2}} ight)={sqrt {pi }}}{displaystyle int _{-infty }^{infty }e^{-z^{2}}\,dz=lim _{R o infty }oint f\,dz=2pi i\,{ ext{res}}left(f,{frac {a}{2}} ight)={sqrt {pi }}}.
0.2 Bearbeiten
{displaystyle int _{0}^{infty }left({frac {1}{e^{x}-1}}-{frac {1}{x\,e^{x}}}
ight)\,dx=gamma }{displaystyle int _{0}^{infty }left({frac {1}{e^{x}-1}}-{frac {1}{x\,e^{x}}}
ight)\,dx=gamma }
Beweis
Für {displaystyle { ext{Re}}(z)>1\,}{displaystyle { ext{Re}}(z)>1\,} gilt {displaystyle int _{0}^{infty }left({frac {x^{z-1}}{e^{x}-1}}-{frac {x^{z-1}}{x\,e^{x}}}
ight)\,dx=Gamma (z)\,left(zeta (z)-{frac {1}{z-1}}
ight) o gamma }{displaystyle int _{0}^{infty }left({frac {x^{z-1}}{e^{x}-1}}-{frac {x^{z-1}}{x\,e^{x}}}
ight)\,dx=Gamma (z)\,left(zeta (z)-{frac {1}{z-1}}
ight) o gamma } für {displaystyle z o 1\,}{displaystyle z o 1\,}.
1.1 Bearbeiten
{displaystyle int _{0}^{infty }x^{z-1}e^{-x}\,dx=Gamma (z)qquad { ext{Re}}(z)>0}{displaystyle int _{0}^{infty }x^{z-1}e^{-x}\,dx=Gamma (z)qquad { ext{Re}}(z)>0}
ohne Beweis
1.2 Bearbeiten
{displaystyle int _{0}^{infty }exp left(-x^{2}-{frac {alpha ^{2}}{x^{2}}}
ight)\,dx={frac {sqrt {pi }}{2}}\,e^{-2alpha }qquad alpha >0}{displaystyle int _{0}^{infty }exp left(-x^{2}-{frac {alpha ^{2}}{x^{2}}}
ight)\,dx={frac {sqrt {pi }}{2}}\,e^{-2alpha }qquad alpha >0}
1. Beweis
{displaystyle 2I=int _{0}^{infty }exp left(-x^{2}-{frac {alpha ^{2}}{x^{2}}}
ight)\,dx=int _{-infty }^{infty }exp left(-left(x-{frac {alpha }{x}}
ight)^{2}-2alpha
ight)\,dx=int _{-infty }^{infty }e^{-left(x-{frac {alpha }{x}}
ight)^{2}}dxcdot e^{-2alpha }}{displaystyle 2I=int _{0}^{infty }exp left(-x^{2}-{frac {alpha ^{2}}{x^{2}}}
ight)\,dx=int _{-infty }^{infty }exp left(-left(x-{frac {alpha }{x}}
ight)^{2}-2alpha
ight)\,dx=int _{-infty }^{infty }e^{-left(x-{frac {alpha }{x}}
ight)^{2}}dxcdot e^{-2alpha }}.
Nach der Formel {displaystyle int _{-infty }^{infty }fleft(x-{frac {b}{x}} ight)dx=int _{-infty }^{infty }f(x)dx}{displaystyle int _{-infty }^{infty }fleft(x-{frac {b}{x}} ight)dx=int _{-infty }^{infty }f(x)dx}, gilt im Fall {displaystyle f(x)=e^{-x^{2}}}{displaystyle f(x)=e^{-x^{2}}},
{displaystyle int _{-infty }^{infty }e^{-left(x-{frac {alpha }{x}} ight)^{2}}dx=int _{-infty }^{infty }e^{-x^{2}}dx}{displaystyle int _{-infty }^{infty }e^{-left(x-{frac {alpha }{x}} ight)^{2}}dx=int _{-infty }^{infty }e^{-x^{2}}dx}, und das ist {displaystyle {sqrt {pi }}}sqrt{pi}.
2. Beweis
{displaystyle I'(alpha )=int _{0}^{infty }exp left(-x^{2}-{frac {alpha ^{2}}{x^{2}}}
ight)\,{frac {-2alpha }{x^{2}}}\,dx}{displaystyle I'(alpha )=int _{0}^{infty }exp left(-x^{2}-{frac {alpha ^{2}}{x^{2}}}
ight)\,{frac {-2alpha }{x^{2}}}\,dx} ist nach Substitution {displaystyle xmapsto {frac {alpha }{x}}}{displaystyle xmapsto {frac {alpha }{x}}} gleich {displaystyle -2int _{0}^{infty }exp left(-x^{2}-{frac {alpha ^{2}}{x^{2}}}
ight)\,dx}{displaystyle -2int _{0}^{infty }exp left(-x^{2}-{frac {alpha ^{2}}{x^{2}}}
ight)\,dx}.
Die Differenzialgleichung {displaystyle I'(alpha )=-2\,I(alpha )}{displaystyle I'(alpha )=-2\,I(alpha )} wird gelöst durch {displaystyle I(alpha )=C\,e^{-2alpha }}{displaystyle I(alpha )=C\,e^{-2alpha }}, wobei {displaystyle C=I(0)=int _{0}^{infty }e^{-x^{2}}\,dx={frac {sqrt {pi }}{2}}}{displaystyle C=I(0)=int _{0}^{infty }e^{-x^{2}}\,dx={frac {sqrt {pi }}{2}}} ist.
1.3 Bearbeiten
{displaystyle int _{0}^{infty }left({frac {1}{x\,e^{x}}}-{frac {e^{-x(z-1)}}{e^{x}-1}}
ight)dx=psi (z)qquad { ext{Re}}(z)>0}{displaystyle int _{0}^{infty }left({frac {1}{x\,e^{x}}}-{frac {e^{-x(z-1)}}{e^{x}-1}}
ight)dx=psi (z)qquad { ext{Re}}(z)>0}
Beweis
In der Formel {displaystyle int _{0}^{1}{frac {1-x^{z-1}}{1-x}}\,dx=psi (z)+gamma }{displaystyle int _{0}^{1}{frac {1-x^{z-1}}{1-x}}\,dx=psi (z)+gamma } wird das Integral
nach Substitution {displaystyle xmapsto e^{-x}}{displaystyle xmapsto e^{-x}} zu {displaystyle int _{0}^{infty }left({frac {1}{e^{x}-1}}-{frac {e^{-x(z-1)}}{e^{x}-1}} ight)\,dx}{displaystyle int _{0}^{infty }left({frac {1}{e^{x}-1}}-{frac {e^{-x(z-1)}}{e^{x}-1}} ight)\,dx},
und {displaystyle gamma \,}{displaystyle gamma \,} lässt sich schreiben als {displaystyle int _{0}^{infty }left({frac {1}{e^{x}-1}}-{frac {1}{x\,e^{x}}} ight)\,dx}{displaystyle int _{0}^{infty }left({frac {1}{e^{x}-1}}-{frac {1}{x\,e^{x}}} ight)\,dx}.
1.4 Bearbeiten
{displaystyle int _{0}^{infty }left({frac {z-1}{x\,e^{x}}}-{frac {1-e^{-x(z-1)}}{x\,(e^{x}-1)}}
ight)\,dx=log Gamma (z)qquad { ext{Re}}(z)>0}{displaystyle int _{0}^{infty }left({frac {z-1}{x\,e^{x}}}-{frac {1-e^{-x(z-1)}}{x\,(e^{x}-1)}}
ight)\,dx=log Gamma (z)qquad { ext{Re}}(z)>0}
Beweis (Formel nach Malmstén)
Integriere die Formel {displaystyle int _{0}^{infty }left({frac {1}{x\,e^{x}}}-{frac {e^{-x(z'-1)}}{e^{x}-1}}
ight)dx=psi (z')}{displaystyle int _{0}^{infty }left({frac {1}{x\,e^{x}}}-{frac {e^{-x(z'-1)}}{e^{x}-1}}
ight)dx=psi (z')} nach z' von 1 bis z.
1.5 Bearbeiten
{displaystyle int _{0}^{infty }left({frac {1}{e^{x}}}-{frac {1}{(1+x)^{z}}}
ight)\,{frac {dx}{x}}=psi (z)qquad { ext{Re}}(z)>0}{displaystyle int _{0}^{infty }left({frac {1}{e^{x}}}-{frac {1}{(1+x)^{z}}}
ight)\,{frac {dx}{x}}=psi (z)qquad { ext{Re}}(z)>0}
Beweis (Formel nach Cauchy)
{displaystyle forall varepsilon >0}{displaystyle forall varepsilon >0} ist {displaystyle int _{0}^{infty }left({frac {x^{varepsilon -1}}{e^{x}}}-{frac {x^{varepsilon -1}}{(1+x)^{z}}}
ight)dx=Gamma (varepsilon )-{frac {Gamma (z-varepsilon )\,Gamma (varepsilon )}{Gamma (z)}}}{displaystyle int _{0}^{infty }left({frac {x^{varepsilon -1}}{e^{x}}}-{frac {x^{varepsilon -1}}{(1+x)^{z}}}
ight)dx=Gamma (varepsilon )-{frac {Gamma (z-varepsilon )\,Gamma (varepsilon )}{Gamma (z)}}}
{displaystyle ={frac {Gamma (z)\,Gamma (varepsilon )-Gamma (z-varepsilon )\,Gamma (varepsilon )}{Gamma (z)}}={frac {Gamma (1+varepsilon )}{Gamma (z)}}cdot {frac {Gamma (z)-Gamma (z-varepsilon )}{varepsilon }}\,{xrightarrow {\,\,varepsilon o 0\,\,}}\,{frac {1}{Gamma (z)}}cdot Gamma '(z)=psi (z)}{displaystyle ={frac {Gamma (z)\,Gamma (varepsilon )-Gamma (z-varepsilon )\,Gamma (varepsilon )}{Gamma (z)}}={frac {Gamma (1+varepsilon )}{Gamma (z)}}cdot {frac {Gamma (z)-Gamma (z-varepsilon )}{varepsilon }}\,{xrightarrow {\,\,varepsilon o 0\,\,}}\,{frac {1}{Gamma (z)}}cdot Gamma '(z)=psi (z)}.
1.6 Bearbeiten
{displaystyle int _{0}^{infty }{frac {x^{z-1}}{e^{x}-1}}\,dx=Gamma (z)\,zeta (z)qquad { ext{Re}}(z)>1}{displaystyle int _{0}^{infty }{frac {x^{z-1}}{e^{x}-1}}\,dx=Gamma (z)\,zeta (z)qquad { ext{Re}}(z)>1}
Beweis
Wegen {displaystyle {frac {1}{e^{x}-1}}={frac {e^{-x}}{1-e^{-x}}}=sum _{k=1}^{infty }e^{-kz}}{displaystyle {frac {1}{e^{x}-1}}={frac {e^{-x}}{1-e^{-x}}}=sum _{k=1}^{infty }e^{-kz}} ist {displaystyle {frac {x^{z-1}}{e^{x}-1}}=sum _{k=1}^{infty }x^{z-1}\,e^{-kx}}{displaystyle {frac {x^{z-1}}{e^{x}-1}}=sum _{k=1}^{infty }x^{z-1}\,e^{-kx}}
und somit {displaystyle int _{0}^{infty }{frac {x^{z-1}}{e^{x}-1}}\,dx=sum _{k=1}^{infty }int _{0}^{infty }x^{z-1}\,e^{-kx}\,dx=sum _{k=1}^{infty }{frac {Gamma (z)}{k^{z}}}=Gamma (z)\,zeta (z)}{displaystyle int _{0}^{infty }{frac {x^{z-1}}{e^{x}-1}}\,dx=sum _{k=1}^{infty }int _{0}^{infty }x^{z-1}\,e^{-kx}\,dx=sum _{k=1}^{infty }{frac {Gamma (z)}{k^{z}}}=Gamma (z)\,zeta (z)}.
1.7 Bearbeiten
{displaystyle int _{0}^{infty }{frac {x^{z-1}}{e^{x}+1}}\,dx=Gamma (z)\,eta (z)qquad { ext{Re}}(z)>0}{displaystyle int _{0}^{infty }{frac {x^{z-1}}{e^{x}+1}}\,dx=Gamma (z)\,eta (z)qquad { ext{Re}}(z)>0}
Beweis
Wegen {displaystyle {frac {1}{e^{x}+1}}={frac {e^{-x}}{1+e^{-x}}}={frac {-(-e^{-x})}{1-(-e^{-x})}}=-sum _{k=1}^{infty }(-e^{-x})^{k}=sum _{k=1}^{infty }(-1)^{k-1}\,e^{-kx}}{displaystyle {frac {1}{e^{x}+1}}={frac {e^{-x}}{1+e^{-x}}}={frac {-(-e^{-x})}{1-(-e^{-x})}}=-sum _{k=1}^{infty }(-e^{-x})^{k}=sum _{k=1}^{infty }(-1)^{k-1}\,e^{-kx}} ist {displaystyle {frac {x^{z-1}}{e^{x}+1}}=sum _{k=1}^{infty }(-1)^{k-1}\,x^{z-1}\,e^{-kx}}{displaystyle {frac {x^{z-1}}{e^{x}+1}}=sum _{k=1}^{infty }(-1)^{k-1}\,x^{z-1}\,e^{-kx}}
und somit {displaystyle int _{0}^{infty }{frac {x^{z-1}}{e^{x}+1}}\,dx=sum _{k=1}^{infty }(-1)^{k-1}\,int _{0}^{infty }x^{z-1}\,e^{-kx}\,dx=sum _{k=1}^{infty }(-1)^{k-1}\,{frac {Gamma (z)}{k^{z}}}=Gamma (z)\,eta (z)}{displaystyle int _{0}^{infty }{frac {x^{z-1}}{e^{x}+1}}\,dx=sum _{k=1}^{infty }(-1)^{k-1}\,int _{0}^{infty }x^{z-1}\,e^{-kx}\,dx=sum _{k=1}^{infty }(-1)^{k-1}\,{frac {Gamma (z)}{k^{z}}}=Gamma (z)\,eta (z)}.
1.8 Bearbeiten
{displaystyle {frac {1}{Gamma (z)}}={frac {1}{2pi i}}int _{C}t^{-z}e^{t}\,dtqquad C\,}{displaystyle {frac {1}{Gamma (z)}}={frac {1}{2pi i}}int _{C}t^{-z}e^{t}\,dtqquad C\,} ist eine Kurve in {displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}}{displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}}, die von {displaystyle -infty -ivarepsilon \,}{displaystyle -infty -ivarepsilon \,} nach {displaystyle -infty +ivarepsilon \,}{displaystyle -infty +ivarepsilon \,} läuft.
Für {displaystyle { ext{Re}}(z)>0\,}{displaystyle { ext{Re}}(z)>0\,} kann man als Integrationsweg {displaystyle C\,}{displaystyle C\,} auch die Gerade {displaystyle a+imathbb {R} }{displaystyle a+imathbb {R} }, mit {displaystyle a>0\,}{displaystyle a>0\,}, hernehmen.
Beweis für Re(z)>0 (Hankelsche Integraldarstellung für die reziproke Gammafunktion)
Die Funktion {displaystyle f(t)=t^{-z}\,e^{t}}{displaystyle f(t)=t^{-z}\,e^{t}} mit {displaystyle { ext{Re}}(z)>0\,}{displaystyle { ext{Re}}(z)>0\,} ist auf {displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}}{displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}} holomorph.
Integrationsweg9.PNG
Für {displaystyle varepsilon leq tleq R\,}{displaystyle varepsilon leq tleq R\,} ist {displaystyle |f(-R+it)|\,}{displaystyle |f(-R+it)|\,}
{displaystyle =left|(-R+it)^{-z} ight|cdot left|e^{-R+it} ight|=Theta left(R^{-{ ext{Re}}(z)} ight)cdot e^{-R}}{displaystyle =left|(-R+it)^{-z} ight|cdot left|e^{-R+it} ight|=Theta left(R^{-{ ext{Re}}(z)} ight)cdot e^{-R}}.
Daher verschwinden die Integrale über den Abschnitten {displaystyle sigma _{1},sigma _{2}\,}{displaystyle sigma _{1},sigma _{2}\,} für {displaystyle R o infty \,}R o infty \,.
Und es ist
{displaystyle left|int _{-R}^{a}f(tpm iR)\,dt ight|leq int _{-R}^{a}left|(tpm iR)^{-z} ight|cdot left|e^{tpm iR} ight|\,dt}{displaystyle left|int _{-R}^{a}f(tpm iR)\,dt ight|leq int _{-R}^{a}left|(tpm iR)^{-z} ight|cdot left|e^{tpm iR} ight|\,dt}
{displaystyle leq max _{-Rleq tleq a}left|(tpm iR)^{-z} ight|cdot int _{-R}^{a}e^{t}\,dt=Theta left(R^{-{ ext{Re}}(z)} ight)cdot e^{a}}{displaystyle leq max _{-Rleq tleq a}left|(tpm iR)^{-z} ight|cdot int _{-R}^{a}e^{t}\,dt=Theta left(R^{-{ ext{Re}}(z)} ight)cdot e^{a}}.
Daher verschwinden auch die Integrale über den Abschnitten {displaystyle au _{1}, au _{2}\,}{displaystyle au _{1}, au _{2}\,} für {displaystyle R o infty \,}R o infty \,.
Also ist
{displaystyle {frac {1}{2pi i}}int _{C}t^{-z}e^{t}\,dt={frac {1}{2pi i}}int _{a-iinfty }^{a+iinfty }s^{-z}\,e^{s}\,ds}{displaystyle {frac {1}{2pi i}}int _{C}t^{-z}e^{t}\,dt={frac {1}{2pi i}}int _{a-iinfty }^{a+iinfty }s^{-z}\,e^{s}\,ds}
{displaystyle ={mathcal {L}}^{-1}left[s^{-z} ight](1)=left.{frac {t^{z-1}}{Gamma (z)}} ight|_{t=1}={frac {1}{Gamma (z)}}}{displaystyle ={mathcal {L}}^{-1}left[s^{-z} ight](1)=left.{frac {t^{z-1}}{Gamma (z)}} ight|_{t=1}={frac {1}{Gamma (z)}}}.
Beweis für Re(z)<1
Die Funktion {displaystyle f(t)=t^{z-1}e^{t}\,}{displaystyle f(t)=t^{z-1}e^{t}\,} mit {displaystyle { ext{Re}}(z)>0\,}{displaystyle { ext{Re}}(z)>0\,} ist auf {displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}}{displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}} holomorph.
Hankelintegrationsweg.PNG
Das Integral über dem Kreisbogen {displaystyle K_{varepsilon }\,}{displaystyle K_{varepsilon }\,} verschwindet für {displaystyle varepsilon o 0+\,}{displaystyle varepsilon o 0+\,}, weil wegen {displaystyle left|t^{z-1} ight|=Theta left(|t|^{{ ext{Re}}(z)-1} ight)=oleft({frac {1}{|t|}} ight)}{displaystyle left|t^{z-1} ight|=Theta left(|t|^{{ ext{Re}}(z)-1} ight)=oleft({frac {1}{|t|}} ight)} für {displaystyle |t| o 0\,}{displaystyle |t| o 0\,}
ist {displaystyle max _{tin K_{varepsilon }}left|t^{z-1}e^{-t} ight|=oleft({frac {1}{varepsilon }} ight)}{displaystyle max _{tin K_{varepsilon }}left|t^{z-1}e^{-t} ight|=oleft({frac {1}{varepsilon }} ight)}, und daher gilt {displaystyle left|int _{K_{varepsilon }}f(t)\,dt ight|leq pi varepsilon cdot oleft({frac {1}{varepsilon }} ight)=o(1)}{displaystyle left|int _{K_{varepsilon }}f(t)\,dt ight|leq pi varepsilon cdot oleft({frac {1}{varepsilon }} ight)=o(1)}.
Für die horizontalen Integrationswege gilt:
{displaystyle int _{-infty }^{0}f(tpm ivarepsilon )\,dt=int _{-infty }^{0}(tpm ivarepsilon )^{z-1}\,e^{tpm ivarepsilon }\,dt=e^{pm ipi (z-1)}int _{-infty }^{0}(-tmp ivarepsilon )^{z-1}\,e^{tpm ivarepsilon }\,dt}{displaystyle int _{-infty }^{0}f(tpm ivarepsilon )\,dt=int _{-infty }^{0}(tpm ivarepsilon )^{z-1}\,e^{tpm ivarepsilon }\,dt=e^{pm ipi (z-1)}int _{-infty }^{0}(-tmp ivarepsilon )^{z-1}\,e^{tpm ivarepsilon }\,dt}
{displaystyle =-e^{pm ipi z}int _{0}^{infty }(tmp ivarepsilon )^{z-1}\,e^{-tpm ivarepsilon }\,dt o -e^{pm ipi z}\,Gamma (z)}{displaystyle =-e^{pm ipi z}int _{0}^{infty }(tmp ivarepsilon )^{z-1}\,e^{-tpm ivarepsilon }\,dt o -e^{pm ipi z}\,Gamma (z)} für {displaystyle varepsilon o 0+\,}{displaystyle varepsilon o 0+\,}.
Daher ist {displaystyle int _{C}t^{z-1}\,e^{t}\,dt=left(e^{ipi z}-e^{-ipi z} ight)Gamma (z)=2isin pi z\,Gamma (z)={frac {2pi i}{Gamma (z)\,Gamma (1-z)}}\,Gamma (z)={frac {2pi i}{Gamma (1-z)}}}{displaystyle int _{C}t^{z-1}\,e^{t}\,dt=left(e^{ipi z}-e^{-ipi z} ight)Gamma (z)=2isin pi z\,Gamma (z)={frac {2pi i}{Gamma (z)\,Gamma (1-z)}}\,Gamma (z)={frac {2pi i}{Gamma (1-z)}}}.
Ersetzt man {displaystyle z\,}z\, durch {displaystyle 1-z\,}{displaystyle 1-z\,}, so ist {displaystyle {frac {1}{Gamma (z)}}={frac {1}{2pi i}}int _{C}t^{-z}e^{t}\,dt}{displaystyle {frac {1}{Gamma (z)}}={frac {1}{2pi i}}int _{C}t^{-z}e^{t}\,dt} für {displaystyle { ext{Re}}(z)<1\,}{displaystyle { ext{Re}}(z)<1\,}.
1.9 Bearbeiten
{displaystyle J_{
u }(x)={frac {1}{2pi i}}int _{C}e^{{frac {x}{2}}left(t-{frac {1}{t}}
ight)}\,{frac {dt}{t^{
u +1}}}qquad { ext{Re}}(x)>0qquad C\,}{displaystyle J_{
u }(x)={frac {1}{2pi i}}int _{C}e^{{frac {x}{2}}left(t-{frac {1}{t}}
ight)}\,{frac {dt}{t^{
u +1}}}qquad { ext{Re}}(x)>0qquad C\,} ist eine Kurve in {displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}}{displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}}, die von {displaystyle -infty -ivarepsilon \,}{displaystyle -infty -ivarepsilon \,} nach {displaystyle -infty +ivarepsilon \,}{displaystyle -infty +ivarepsilon \,} läuft.
Beweis für x>0 (Hankelsche Integraldarstellung für die Besselfunktion)
{displaystyle J_{
u }(x)=sum _{n=0}^{infty }{frac {(-1)^{n}}{n!\,Gamma (n+
u +1)}}left({frac {x}{2}}
ight)^{
u +2n}}{displaystyle J_{
u }(x)=sum _{n=0}^{infty }{frac {(-1)^{n}}{n!\,Gamma (n+
u +1)}}left({frac {x}{2}}
ight)^{
u +2n}}
Ersetze {displaystyle {frac {1}{Gamma (n+ u +1)}}}{displaystyle {frac {1}{Gamma (n+ u +1)}}} durch die Hankelsche Integraldarstellung {displaystyle {frac {1}{2pi i}}int _{C}t^{-n- u -1}e^{t}\,dt}{displaystyle {frac {1}{2pi i}}int _{C}t^{-n- u -1}e^{t}\,dt}.
{displaystyle J_{ u }(x)={frac {1}{2pi i}}int _{C}sum _{n=0}^{infty }{frac {(-1)^{n}}{n!}}\,{frac {left({frac {x}{2}} ight)^{2n}}{t^{n}}}\,left({frac {x}{2}} ight)^{ u }\,{frac {e^{t}}{t^{ u +1}}}\,dt={frac {1}{2pi i}}int _{C}e^{t-{frac {x^{2}}{4t}}}\,left({frac {x}{2}} ight)^{ u }{frac {dt}{t^{ u +1}}}}{displaystyle J_{ u }(x)={frac {1}{2pi i}}int _{C}sum _{n=0}^{infty }{frac {(-1)^{n}}{n!}}\,{frac {left({frac {x}{2}} ight)^{2n}}{t^{n}}}\,left({frac {x}{2}} ight)^{ u }\,{frac {e^{t}}{t^{ u +1}}}\,dt={frac {1}{2pi i}}int _{C}e^{t-{frac {x^{2}}{4t}}}\,left({frac {x}{2}} ight)^{ u }{frac {dt}{t^{ u +1}}}}
Nach Substitution {displaystyle tmapsto {frac {x}{2}}cdot t}{displaystyle tmapsto {frac {x}{2}}cdot t} ändert sich am Integrationsweg {displaystyle C\,}{displaystyle C\,} nichts, und es ist {displaystyle J_{ u }(x)={frac {1}{2pi i}}int _{C}e^{{frac {x}{2}}left(t-{frac {1}{t}} ight)}\,{frac {dt}{t^{ u +1}}}}{displaystyle J_{ u }(x)={frac {1}{2pi i}}int _{C}e^{{frac {x}{2}}left(t-{frac {1}{t}} ight)}\,{frac {dt}{t^{ u +1}}}}.
1.10 Bearbeiten
{displaystyle int _{0}^{infty }left({frac {1}{x}}-{frac {1}{e^{x}-1}}
ight)e^{-zx}\,dx=psi (z+1)-log zqquad { ext{Re}}(z)>0}{displaystyle int _{0}^{infty }left({frac {1}{x}}-{frac {1}{e^{x}-1}}
ight)e^{-zx}\,dx=psi (z+1)-log zqquad { ext{Re}}(z)>0}
Beweis
Dies folgt unmittelbar aus den Formeln {displaystyle psi (z+1)=int _{0}^{infty }left({frac {e^{-x}}{x}}-{frac {e^{-zx}}{e^{x}-1}}
ight)dx}{displaystyle psi (z+1)=int _{0}^{infty }left({frac {e^{-x}}{x}}-{frac {e^{-zx}}{e^{x}-1}}
ight)dx} und {displaystyle log z=int _{0}^{infty }left({frac {e^{-x}}{x}}-{frac {e^{-zx}}{x}}
ight)dx}{displaystyle log z=int _{0}^{infty }left({frac {e^{-x}}{x}}-{frac {e^{-zx}}{x}}
ight)dx}.
1.11 Bearbeiten
{displaystyle int _{0}^{infty }left({frac {1}{2}}-{frac {1}{x}}+{frac {1}{e^{x}-1}}
ight)\,{frac {e^{-zx}}{x}}\,dx=log left({frac {z!\,e^{z}}{z^{z}\,{sqrt {2pi z}}}}
ight)qquad { ext{Re}}(z)>0}{displaystyle int _{0}^{infty }left({frac {1}{2}}-{frac {1}{x}}+{frac {1}{e^{x}-1}}
ight)\,{frac {e^{-zx}}{x}}\,dx=log left({frac {z!\,e^{z}}{z^{z}\,{sqrt {2pi z}}}}
ight)qquad { ext{Re}}(z)>0}
Beweis (Erste Binetsche Formel)
In der Formel {displaystyle psi (z)+{frac {1}{z}}=log z+int _{0}^{infty }e^{-zx}left({frac {1}{x}}-{frac {1}{e^{x}-1}}
ight)dx}{displaystyle psi (z)+{frac {1}{z}}=log z+int _{0}^{infty }e^{-zx}left({frac {1}{x}}-{frac {1}{e^{x}-1}}
ight)dx} ersetze {displaystyle {frac {1}{z}}}{displaystyle {frac {1}{z}}} durch {displaystyle int _{0}^{infty }e^{-zx}\,dx}{displaystyle int _{0}^{infty }e^{-zx}\,dx}:
{displaystyle psi (z)=log z-{frac {1}{2z}}+int _{0}^{infty }e^{-zx}left(-{frac {1}{2}}+{frac {1}{x}}-{frac {1}{e^{x}-1}} ight)dx}{displaystyle psi (z)=log z-{frac {1}{2z}}+int _{0}^{infty }e^{-zx}left(-{frac {1}{2}}+{frac {1}{x}}-{frac {1}{e^{x}-1}} ight)dx}, wobei {displaystyle lim _{x o 0}{frac {{frac {1}{2}}-{frac {1}{x}}+{frac {1}{e^{x}-1}}}{x}}={frac {1}{12}}}{displaystyle lim _{x o 0}{frac {{frac {1}{2}}-{frac {1}{x}}+{frac {1}{e^{x}-1}}}{x}}={frac {1}{12}}} ist.
Integriert man beide Seiten unbestimmt nach {displaystyle z\,}z\,, so ist
{displaystyle log Gamma (z)=left(z-{frac {1}{2}} ight)log z-z+int _{0}^{infty }e^{-zx}left({frac {1}{2}}-{frac {1}{x}}+{frac {1}{e^{x}-1}} ight){frac {dx}{x}}+C}{displaystyle log Gamma (z)=left(z-{frac {1}{2}} ight)log z-z+int _{0}^{infty }e^{-zx}left({frac {1}{2}}-{frac {1}{x}}+{frac {1}{e^{x}-1}} ight){frac {dx}{x}}+C}.
Daraus folgt {displaystyle int _{0}^{infty }e^{-zx}left({frac {1}{2}}-{frac {1}{x}}+{frac {1}{e^{x}-1}} ight){frac {dx}{x}}+C=log left({frac {(z-1)!\,e^{z}}{z^{z-{frac {1}{2}}}}} ight)}{displaystyle int _{0}^{infty }e^{-zx}left({frac {1}{2}}-{frac {1}{x}}+{frac {1}{e^{x}-1}} ight){frac {dx}{x}}+C=log left({frac {(z-1)!\,e^{z}}{z^{z-{frac {1}{2}}}}} ight)}.
Nachdem für {displaystyle z o infty \,}{displaystyle z o infty \,} das Integral verschwindet, ist {displaystyle C=lim _{z o infty }log left({frac {z!\,e^{z}}{z^{z}\,{sqrt {z}}}} ight)=log {sqrt {2pi }}}{displaystyle C=lim _{z o infty }log left({frac {z!\,e^{z}}{z^{z}\,{sqrt {z}}}} ight)=log {sqrt {2pi }}}.
1.12 Bearbeiten
{displaystyle J_{
u }(z)={frac {1}{Gamma left({frac {1}{2}}
ight)Gamma left(
u +{frac {1}{2}}
ight)}}left({frac {z}{2}}
ight)^{
u }int _{-1}^{1}e^{izx}\,(1-x^{2})^{
u -{frac {1}{2}}}\,dxqquad { ext{Re}}(
u )>-{frac {1}{2}}}{displaystyle J_{
u }(z)={frac {1}{Gamma left({frac {1}{2}}
ight)Gamma left(
u +{frac {1}{2}}
ight)}}left({frac {z}{2}}
ight)^{
u }int _{-1}^{1}e^{izx}\,(1-x^{2})^{
u -{frac {1}{2}}}\,dxqquad { ext{Re}}(
u )>-{frac {1}{2}}}
Beweis (Poissonsche Darstellung der Besselfunktion)
Setze {displaystyle I:=int _{-1}^{1}e^{izx}\,(1-x^{2})^{
u -{frac {1}{2}}}\,dx}{displaystyle I:=int _{-1}^{1}e^{izx}\,(1-x^{2})^{
u -{frac {1}{2}}}\,dx}.
Verwende die Reihenentwicklung {displaystyle e^{izx}=sum _{n=0}^{infty }{frac {(iz)^{n}}{n!}}\,x^{n}}{displaystyle e^{izx}=sum _{n=0}^{infty }{frac {(iz)^{n}}{n!}}\,x^{n}}:
{displaystyle I=sum _{n=0}^{infty }{frac {(iz)^{n}}{n!}}int _{-1}^{1}x^{n}(1-x^{2})^{ u -{frac {1}{2}}}\,dx}{displaystyle I=sum _{n=0}^{infty }{frac {(iz)^{n}}{n!}}int _{-1}^{1}x^{n}(1-x^{2})^{ u -{frac {1}{2}}}\,dx}
Letzter Integrand ist für gerade {displaystyle n\,}n\, gerade und für ungerade {displaystyle n\,}n\, ungerade.
{displaystyle I=sum _{n=0}^{infty }{frac {(iz)^{2n}}{(2n)!}}cdot 2int _{0}^{1}x^{2n}\,(1-x^{2})^{ u -{frac {1}{2}}}\,dx}{displaystyle I=sum _{n=0}^{infty }{frac {(iz)^{2n}}{(2n)!}}cdot 2int _{0}^{1}x^{2n}\,(1-x^{2})^{ u -{frac {1}{2}}}\,dx}
Nach Substitution {displaystyle x={sqrt {t}}}{displaystyle x={sqrt {t}}} ist {displaystyle I=sum _{n=0}^{infty }{frac {(-1)^{n}\,z^{2n}}{(2n)!}}\,int _{0}^{1}t^{n-{frac {1}{2}}}\,(1-t)^{ u -{frac {1}{2}}}\,dt}{displaystyle I=sum _{n=0}^{infty }{frac {(-1)^{n}\,z^{2n}}{(2n)!}}\,int _{0}^{1}t^{n-{frac {1}{2}}}\,(1-t)^{ u -{frac {1}{2}}}\,dt}.
Dabei ist {displaystyle int _{0}^{1}t^{n-{frac {1}{2}}}\,(1-t)^{ u -{frac {1}{2}}}\,dt=Bleft(n+{frac {1}{2}}, u +{frac {1}{2}} ight)}{displaystyle int _{0}^{1}t^{n-{frac {1}{2}}}\,(1-t)^{ u -{frac {1}{2}}}\,dt=Bleft(n+{frac {1}{2}}, u +{frac {1}{2}} ight)} {displaystyle ={frac {Gamma left(n+{frac {1}{2}} ight)Gamma left( u +{frac {1}{2}} ight)}{Gamma ( u +n+1)}}}{displaystyle ={frac {Gamma left(n+{frac {1}{2}} ight)Gamma left( u +{frac {1}{2}} ight)}{Gamma ( u +n+1)}}},
wobei nach Legendrescher Verdopplungsformel {displaystyle Gamma left(n+{frac {1}{2}} ight)={frac {Gamma left({frac {1}{2}} ight)}{2^{2n}}}\,{frac {(2n)!}{n!}}}{displaystyle Gamma left(n+{frac {1}{2}} ight)={frac {Gamma left({frac {1}{2}} ight)}{2^{2n}}}\,{frac {(2n)!}{n!}}} ist.
Also ist {displaystyle I=sum _{n=0}^{infty }(-1)^{n}\,left({frac {z}{2}} ight)^{2n}\,{frac {Gamma left({frac {1}{2}} ight)\,Gamma left( u +{frac {1}{2}} ight)}{n!\,Gamma ( u +n+1)}}}{displaystyle I=sum _{n=0}^{infty }(-1)^{n}\,left({frac {z}{2}} ight)^{2n}\,{frac {Gamma left({frac {1}{2}} ight)\,Gamma left( u +{frac {1}{2}} ight)}{n!\,Gamma ( u +n+1)}}},
und damit ist {displaystyle {frac {1}{Gamma left({frac {1}{2}} ight)Gamma left( u +{frac {1}{2}} ight)}}left({frac {z}{2}} ight)^{ u }I=sum _{n=0}^{infty }{frac {(-1)^{n}}{n!\,Gamma ( u +n+1)}}left({frac {z}{2}} ight)^{ u +2n}=J_{ u }(z)}{displaystyle {frac {1}{Gamma left({frac {1}{2}} ight)Gamma left( u +{frac {1}{2}} ight)}}left({frac {z}{2}} ight)^{ u }I=sum _{n=0}^{infty }{frac {(-1)^{n}}{n!\,Gamma ( u +n+1)}}left({frac {z}{2}} ight)^{ u +2n}=J_{ u }(z)}.
2.1 Bearbeiten
{displaystyle int _{0}^{infty }e^{-sx}\,{frac {a}{a^{2}+x^{2}}}\,dx=sin(as)\,{ ext{Ci}}(as)-cos(as)\,left({ ext{Si}}(as)-{frac {pi }{2}}
ight)}{displaystyle int _{0}^{infty }e^{-sx}\,{frac {a}{a^{2}+x^{2}}}\,dx=sin(as)\,{ ext{Ci}}(as)-cos(as)\,left({ ext{Si}}(as)-{frac {pi }{2}}
ight)}
Beweis
{displaystyle int _{0}^{infty }e^{-sx}\,{frac {a}{a^{2}+x^{2}}}\,dx=int _{0}^{infty }e^{-sx}int _{0}^{infty }sin(at)\,e^{-xt}\,dt\,dx=int _{0}^{infty }int _{0}^{infty }sin(at)\,e^{-(s+t)x}\,dx\,dt}{displaystyle int _{0}^{infty }e^{-sx}\,{frac {a}{a^{2}+x^{2}}}\,dx=int _{0}^{infty }e^{-sx}int _{0}^{infty }sin(at)\,e^{-xt}\,dt\,dx=int _{0}^{infty }int _{0}^{infty }sin(at)\,e^{-(s+t)x}\,dx\,dt}
{displaystyle =int _{0}^{infty }{frac {sin(at)}{s+t}}\,dt=int _{s}^{infty }{frac {sin left(a(t-s) ight)}{t}}\,dt=int _{s}^{infty }{frac {sin(at)\,cos(as)-cos(at)\,sin(as)}{t}}\,dt}{displaystyle =int _{0}^{infty }{frac {sin(at)}{s+t}}\,dt=int _{s}^{infty }{frac {sin left(a(t-s) ight)}{t}}\,dt=int _{s}^{infty }{frac {sin(at)\,cos(as)-cos(at)\,sin(as)}{t}}\,dt}
{displaystyle =cos(as)int _{s}^{infty }{frac {sin(at)}{t}}\,dt-sin(as)int _{s}^{infty }{frac {cos(at)}{t}}\,dt=cos(as)\,left({frac {pi }{2}}-{ ext{Si}}(as) ight)+sin(as)\,{ ext{Ci}}(as)}{displaystyle =cos(as)int _{s}^{infty }{frac {sin(at)}{t}}\,dt-sin(as)int _{s}^{infty }{frac {cos(at)}{t}}\,dt=cos(as)\,left({frac {pi }{2}}-{ ext{Si}}(as) ight)+sin(as)\,{ ext{Ci}}(as)}
2.2 Bearbeiten
{displaystyle int _{0}^{infty }e^{-sx}\,{frac {x}{a^{2}+x^{2}}}\,dx=sin(as)\,left({frac {pi }{2}}-{ ext{Si}}(as)
ight)-cos(as)\,{ ext{Ci}}(as)}{displaystyle int _{0}^{infty }e^{-sx}\,{frac {x}{a^{2}+x^{2}}}\,dx=sin(as)\,left({frac {pi }{2}}-{ ext{Si}}(as)
ight)-cos(as)\,{ ext{Ci}}(as)}
Beweis
{displaystyle int _{0}^{infty }e^{-sx}\,{frac {x}{a^{2}+x^{2}}}\,dx=int _{0}^{infty }e^{-sx}int _{0}^{infty }cos(at)\,e^{-xt}\,dt\,dx=int _{0}^{infty }int _{0}^{infty }cos(at)\,e^{-(s+t)x}\,dx\,dt}{displaystyle int _{0}^{infty }e^{-sx}\,{frac {x}{a^{2}+x^{2}}}\,dx=int _{0}^{infty }e^{-sx}int _{0}^{infty }cos(at)\,e^{-xt}\,dt\,dx=int _{0}^{infty }int _{0}^{infty }cos(at)\,e^{-(s+t)x}\,dx\,dt}
{displaystyle =int _{0}^{infty }{frac {cos(at)}{s+t}}\,dt=int _{s}^{infty }{frac {cos left(a(t-s) ight)}{t}}\,dt=int _{s}^{infty }{frac {cos(at)\,cos(as)+sin(at)\,sin(as)}{t}}\,dt}{displaystyle =int _{0}^{infty }{frac {cos(at)}{s+t}}\,dt=int _{s}^{infty }{frac {cos left(a(t-s) ight)}{t}}\,dt=int _{s}^{infty }{frac {cos(at)\,cos(as)+sin(at)\,sin(as)}{t}}\,dt}
{displaystyle =cos(as)int _{s}^{infty }{frac {cos(at)}{t}}\,dt+sin(as)int _{s}^{infty }{frac {sin(at)}{t}}\,dt=-cos(as)\,{ ext{Ci}}(as)+sin(as)\,left({frac {pi }{2}}-{ ext{Si}}(as) ight)}{displaystyle =cos(as)int _{s}^{infty }{frac {cos(at)}{t}}\,dt+sin(as)int _{s}^{infty }{frac {sin(at)}{t}}\,dt=-cos(as)\,{ ext{Ci}}(as)+sin(as)\,left({frac {pi }{2}}-{ ext{Si}}(as) ight)}
2.3 Bearbeiten
{displaystyle int _{0}^{infty }x^{z-1}e^{-mu x}\,dx={frac {Gamma (z)}{mu ^{z}}}qquad { ext{Re}}(z),{ ext{Re}}(mu )>0}{displaystyle int _{0}^{infty }x^{z-1}e^{-mu x}\,dx={frac {Gamma (z)}{mu ^{z}}}qquad { ext{Re}}(z),{ ext{Re}}(mu )>0}
ohne Beweis
2.4 Bearbeiten
{displaystyle int _{0}^{infty }x^{z-1}\,e^{-imu x}\,dx={frac {Gamma (z)}{(imu )^{z}}}qquad 0<operatorname {Re} (z)<1\,,\,mu >0}{displaystyle int _{0}^{infty }x^{z-1}\,e^{-imu x}\,dx={frac {Gamma (z)}{(imu )^{z}}}qquad 0<operatorname {Re} (z)<1\,,\,mu >0}
ohne Beweis
2.5 Bearbeiten
{displaystyle int _{-infty }^{infty }{frac {a^{2}}{(e^{x}-ax-b)^{2}+(api )^{2}}}\,dx={frac {1}{1+Wleft({frac {1}{a}}\,e^{-b/a}
ight)}}qquad a>0\,,\,bin mathbb {R} }{displaystyle int _{-infty }^{infty }{frac {a^{2}}{(e^{x}-ax-b)^{2}+(api )^{2}}}\,dx={frac {1}{1+Wleft({frac {1}{a}}\,e^{-b/a}
ight)}}qquad a>0\,,\,bin mathbb {R} }
Beweis
Betrachte die Funktion {displaystyle f(z)={frac {1}{alog(-z)+b-z}}cdot {frac {1}{z}}}{displaystyle f(z)={frac {1}{alog(-z)+b-z}}cdot {frac {1}{z}}} auf dem Gebiet {displaystyle D:=mathbb {C} setminus mathbb {R} ^{geq 0}}{displaystyle D:=mathbb {C} setminus mathbb {R} ^{geq 0}}.
Eins durch Log.PNG
{displaystyle forall zin D}{displaystyle forall zin D} gibt es genau ein {displaystyle r>0\,}{displaystyle r>0\,} und genau ein {displaystyle -{frac {pi }{2}}<varphi <{frac {pi }{2}}}{displaystyle -{frac {pi }{2}}<varphi <{frac {pi }{2}}}, so dass {displaystyle -z=re^{ivarphi }\,}{displaystyle -z=re^{ivarphi }\,} ist.
Beim Nenner {displaystyle alog(-z)+b-z=alog left(re^{ivarphi } ight)+b+re^{ivarphi }}{displaystyle alog(-z)+b-z=alog left(re^{ivarphi } ight)+b+re^{ivarphi }}
{displaystyle =alog r+iavarphi +b+rcos varphi +irsin varphi =(alog r+rcos varphi +b)+i(avarphi +rsin varphi )\,}{displaystyle =alog r+iavarphi +b+rcos varphi +irsin varphi =(alog r+rcos varphi +b)+i(avarphi +rsin varphi )\,}
hat der Imaginärteil das selbe Vorzeichen wie {displaystyle varphi \,}varphi \, und der Realteil steigt streng monoton in {displaystyle r\,}r\,.
Daher ist {displaystyle z=-acdot Wleft({frac {1}{a}}\,e^{-b/a} ight)}{displaystyle z=-acdot Wleft({frac {1}{a}}\,e^{-b/a} ight)} die einzige Polstelle von {displaystyle f\,}f\,.
Diese erhält man, wenn man {displaystyle varphi =0\,}{displaystyle varphi =0\,} und {displaystyle r=acdot Wleft({frac {1}{a}}\,e^{-b/a} ight)}{displaystyle r=acdot Wleft({frac {1}{a}}\,e^{-b/a} ight)} setzt.
Nun ist {displaystyle { ext{res}}left(f,-acdot Wleft({frac {1}{a}}\,e^{-b/a} ight) ight)={frac {1}{a}}cdot {frac {1}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}{displaystyle { ext{res}}left(f,-acdot Wleft({frac {1}{a}}\,e^{-b/a} ight) ight)={frac {1}{a}}cdot {frac {1}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}.
Also gilt nach dem Residuensatz {displaystyle int _{gamma _{R,varepsilon }}f\,dz+int _{C_{R}}f\,dz+int _{delta _{R,varepsilon }}f\,dz+int _{c_{varepsilon }}f\,dz={frac {1}{a}}cdot {frac {2pi i}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}{displaystyle int _{gamma _{R,varepsilon }}f\,dz+int _{C_{R}}f\,dz+int _{delta _{R,varepsilon }}f\,dz+int _{c_{varepsilon }}f\,dz={frac {1}{a}}cdot {frac {2pi i}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}.
Aus {displaystyle L(C_{R})sim 2pi R\,}{displaystyle L(C_{R})sim 2pi R\,} und {displaystyle M(C_{R})=max _{zin C_{R}}|f(z)|sim {frac {1}{R^{2}}}}{displaystyle M(C_{R})=max _{zin C_{R}}|f(z)|sim {frac {1}{R^{2}}}} folgt {displaystyle left|int _{C_{R}}f\,dz ight|leq L(C_{R})\,M(C_{R})sim {frac {2pi }{R}}}{displaystyle left|int _{C_{R}}f\,dz ight|leq L(C_{R})\,M(C_{R})sim {frac {2pi }{R}}}.
Daher geht {displaystyle int _{C_{R}}f\,dz}{displaystyle int _{C_{R}}f\,dz} gegen null für {displaystyle R o infty \,}R o infty \,.
Und aus {displaystyle L(c_{varepsilon })=pi varepsilon \,}{displaystyle L(c_{varepsilon })=pi varepsilon \,} und {displaystyle M(c_{varepsilon })=max _{zin c_{varepsilon }}|f(z)|sim {frac {-1}{a\,log varepsilon }}cdot {frac {1}{varepsilon }}}{displaystyle M(c_{varepsilon })=max _{zin c_{varepsilon }}|f(z)|sim {frac {-1}{a\,log varepsilon }}cdot {frac {1}{varepsilon }}} folgt {displaystyle left|int _{c_{varepsilon }}f\,dz ight|leq L(c_{varepsilon })\,M(c_{varepsilon })sim {frac {-pi }{a\,log varepsilon }}}{displaystyle left|int _{c_{varepsilon }}f\,dz ight|leq L(c_{varepsilon })\,M(c_{varepsilon })sim {frac {-pi }{a\,log varepsilon }}}.
Daher geht {displaystyle int _{c_{varepsilon }}f\,dz}{displaystyle int _{c_{varepsilon }}f\,dz} auch gegen null für {displaystyle varepsilon o 0+\,}{displaystyle varepsilon o 0+\,}.
Im Grenzübergang {displaystyle R o infty \,,\,varepsilon o 0+}{displaystyle R o infty \,,\,varepsilon o 0+} ergibt sich
{displaystyle int _{0}^{infty }left({frac {1}{alog(-x-i0^{+})+b-x}}-{frac {1}{alog(-x+i0^{+})+b-x}} ight){frac {dx}{x}}={frac {1}{a}}cdot {frac {2pi i}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}{displaystyle int _{0}^{infty }left({frac {1}{alog(-x-i0^{+})+b-x}}-{frac {1}{alog(-x+i0^{+})+b-x}} ight){frac {dx}{x}}={frac {1}{a}}cdot {frac {2pi i}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}.
Dabei ist {displaystyle {frac {1}{a(log x-ipi )+b-x}}-{frac {1}{a(log x+ipi )+b-x}}={frac {2pi icdot a}{(alog x+b-x)^{2}-(ipi a)^{2}}}}{displaystyle {frac {1}{a(log x-ipi )+b-x}}-{frac {1}{a(log x+ipi )+b-x}}={frac {2pi icdot a}{(alog x+b-x)^{2}-(ipi a)^{2}}}},
und somit gilt {displaystyle int _{0}^{infty }{frac {a^{2}}{(alog x+b-x)^{2}+(api )^{2}}}cdot {frac {dx}{x}}={frac {1}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}{displaystyle int _{0}^{infty }{frac {a^{2}}{(alog x+b-x)^{2}+(api )^{2}}}cdot {frac {dx}{x}}={frac {1}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}.
Substituiert man {displaystyle xmapsto e^{x}}{displaystyle xmapsto e^{x}}, so ist {displaystyle int _{-infty }^{infty }{frac {a^{2}}{(ax+b-e^{x})^{2}+(api )^{2}}}\,dx={frac {1}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}{displaystyle int _{-infty }^{infty }{frac {a^{2}}{(ax+b-e^{x})^{2}+(api )^{2}}}\,dx={frac {1}{1+Wleft({frac {1}{a}}\,e^{-b/a} ight)}}}.
4.1 Bearbeiten
{displaystyle int _{0}^{infty }{frac {e^{kx}-e^{lambda x}}{e^{mu x}-e^{
u x}}}\,dx={frac {1}{mu -
u }}left(psi left({frac {mu -lambda }{mu -
u }}
ight)-psi left({frac {mu -k}{mu -
u }}
ight)
ight)}{displaystyle int _{0}^{infty }{frac {e^{kx}-e^{lambda x}}{e^{mu x}-e^{
u x}}}\,dx={frac {1}{mu -
u }}left(psi left({frac {mu -lambda }{mu -
u }}
ight)-psi left({frac {mu -k}{mu -
u }}
ight)
ight)}
Beweis
Aus der Gaußschen Formel {displaystyle psi (z)+gamma =int _{0}^{1}{frac {1-u^{z-1}}{1-u}}\,du}{displaystyle psi (z)+gamma =int _{0}^{1}{frac {1-u^{z-1}}{1-u}}\,du}
folgt {displaystyle psi (1-eta )-psi (1-alpha )=int _{0}^{1}{frac {u^{-alpha }-u^{-eta }}{1-u}}\,du=int _{0}^{infty }{frac {e^{alpha x}-e^{eta x}}{e^{x}-1}}\,dxquad left({ ext{nach Substitution}}\,u=e^{-x} ight)}{displaystyle psi (1-eta )-psi (1-alpha )=int _{0}^{1}{frac {u^{-alpha }-u^{-eta }}{1-u}}\,du=int _{0}^{infty }{frac {e^{alpha x}-e^{eta x}}{e^{x}-1}}\,dxquad left({ ext{nach Substitution}}\,u=e^{-x} ight)}.
Nun ist {displaystyle int _{0}^{infty }{frac {e^{kx}-e^{lambda x}}{e^{mu x}-e^{ u x}}}\,dx=int _{0}^{infty }{frac {e^{(k- u )x}-e^{(lambda - u )x}}{e^{(mu - u )x}-1}}\,dx={frac {1}{mu - u }}int _{0}^{infty }{frac {e^{{frac {k- u }{mu - u }}\,x}-e^{{frac {lambda - u }{mu - u }}\,x}}{e^{x}-1}}\,dx}{displaystyle int _{0}^{infty }{frac {e^{kx}-e^{lambda x}}{e^{mu x}-e^{ u x}}}\,dx=int _{0}^{infty }{frac {e^{(k- u )x}-e^{(lambda - u )x}}{e^{(mu - u )x}-1}}\,dx={frac {1}{mu - u }}int _{0}^{infty }{frac {e^{{frac {k- u }{mu - u }}\,x}-e^{{frac {lambda - u }{mu - u }}\,x}}{e^{x}-1}}\,dx}
{displaystyle ={frac {1}{mu - u }}left(psi left(1-{frac {lambda - u }{mu - u }} ight)-psi left(1-{frac {k- u }{mu - u }} ight) ight)={frac {1}{mu - u }}left(psi left({frac {mu -lambda }{mu - u }} ight)-psi left({frac {mu -k}{mu - u }} ight) ight)}{displaystyle ={frac {1}{mu - u }}left(psi left(1-{frac {lambda - u }{mu - u }} ight)-psi left(1-{frac {k- u }{mu - u }} ight) ight)={frac {1}{mu - u }}left(psi left({frac {mu -lambda }{mu - u }} ight)-psi left({frac {mu -k}{mu - u }} ight) ight)}.
Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,log,tan)
0.1 Bearbeiten
{displaystyle int _{0}^{1}log left( an {frac {pi x}{2}}
ight)\,dx=0}{displaystyle int _{0}^{1}log left( an {frac {pi x}{2}}
ight)\,dx=0}
ohne Beweis
0.2 Bearbeiten
{displaystyle int _{0}^{pi }log ^{2}left( an {frac {x}{2}}
ight)dx={frac {pi ^{3}}{4}}}{displaystyle int _{0}^{pi }log ^{2}left( an {frac {x}{2}}
ight)dx={frac {pi ^{3}}{4}}}
Beweis
Die Funktion {displaystyle f(x)=-{frac {1}{2}}log left( an {frac {x}{2}}
ight)}{displaystyle f(x)=-{frac {1}{2}}log left( an {frac {x}{2}}
ight)} besitzt die Fourierreihenentwicklung {displaystyle sum _{k=0}^{infty }{frac {cos(2k+1)x}{2k+1}}}{displaystyle sum _{k=0}^{infty }{frac {cos(2k+1)x}{2k+1}}}.
Nach der Parsevalschen Gleichung {displaystyle {frac {1}{pi }}int _{-pi }^{pi }|f(x)|^{2}\,dx={frac {a_{0}^{2}}{2}}+sum _{k=1}^{infty }{Big (}a_{k}^{2}+b_{k}^{2}{Big )}}{displaystyle {frac {1}{pi }}int _{-pi }^{pi }|f(x)|^{2}\,dx={frac {a_{0}^{2}}{2}}+sum _{k=1}^{infty }{Big (}a_{k}^{2}+b_{k}^{2}{Big )}}
gilt dann {displaystyle {frac {1}{4pi }}int _{-pi }^{pi }log ^{2}left( an {frac {x}{2}} ight)dx=sum _{k=0}^{infty }{frac {1}{(2k+1)^{2}}}={frac {pi ^{2}}{8}}}{displaystyle {frac {1}{4pi }}int _{-pi }^{pi }log ^{2}left( an {frac {x}{2}} ight)dx=sum _{k=0}^{infty }{frac {1}{(2k+1)^{2}}}={frac {pi ^{2}}{8}}}.
0.3 Bearbeiten
{displaystyle int _{0}^{pi }log ^{2}left( an {frac {x}{4}}
ight)dx={frac {pi ^{3}}{4}}}{displaystyle int _{0}^{pi }log ^{2}left( an {frac {x}{4}}
ight)dx={frac {pi ^{3}}{4}}}
Beweis
Die Funktion {displaystyle f(x)=log ^{2}left( an {frac {x}{2}}
ight)}{displaystyle f(x)=log ^{2}left( an {frac {x}{2}}
ight)} besitzt die Symmetrie {displaystyle f(pi -x)=f(x)\,}{displaystyle f(pi -x)=f(x)\,}.
{displaystyle int _{0}^{pi }fleft({frac {x}{2}} ight)dx=2int _{0}^{frac {pi }{2}}f(x)\,dx}{displaystyle int _{0}^{pi }fleft({frac {x}{2}} ight)dx=2int _{0}^{frac {pi }{2}}f(x)\,dx} ist daher {displaystyle int _{0}^{pi }f(x)\,dx={frac {pi ^{3}}{4}}}{displaystyle int _{0}^{pi }f(x)\,dx={frac {pi ^{3}}{4}}}.
0.4 Bearbeiten
{displaystyle int _{pi /4}^{pi /2}log log an x\,dx={frac {pi }{2}}\,log left({sqrt {2pi }}\,\,{frac {Gamma left({frac {3}{4}}
ight)}{Gamma left({frac {1}{4}}
ight)}}
ight)}{displaystyle int _{pi /4}^{pi /2}log log an x\,dx={frac {pi }{2}}\,log left({sqrt {2pi }}\,\,{frac {Gamma left({frac {3}{4}}
ight)}{Gamma left({frac {1}{4}}
ight)}}
ight)}
1. Beweis (Vardisches Integral)
{displaystyle I:=int _{pi /4}^{pi /2}log log an x\,dx}{displaystyle I:=int _{pi /4}^{pi /2}log log an x\,dx} ist nach Substitution {displaystyle xmapsto arctan e^{x}}{displaystyle xmapsto arctan e^{x}} gleich {displaystyle {frac {1}{2}}int _{0}^{infty }{frac {log x}{cosh x}}\,dx}{displaystyle {frac {1}{2}}int _{0}^{infty }{frac {log x}{cosh x}}\,dx}.
Und das ist {displaystyle {frac {pi }{2}}\,int _{0}^{infty }{frac {log pi x}{cosh pi x}}\,dx={frac {pi }{2}}log {sqrt {pi }}int _{-infty }^{infty }{frac {dx}{cosh pi x}}+{frac {pi }{8}}int _{-infty }^{infty }{frac {log x^{2}}{cosh pi x}}\,dx}{displaystyle {frac {pi }{2}}\,int _{0}^{infty }{frac {log pi x}{cosh pi x}}\,dx={frac {pi }{2}}log {sqrt {pi }}int _{-infty }^{infty }{frac {dx}{cosh pi x}}+{frac {pi }{8}}int _{-infty }^{infty }{frac {log x^{2}}{cosh pi x}}\,dx}.
Dabei ist {displaystyle int _{-infty }^{infty }{frac {dx}{cosh pi x}}=1}{displaystyle int _{-infty }^{infty }{frac {dx}{cosh pi x}}=1} und nach der Formel {displaystyle int _{-infty }^{infty }{frac {log(alpha ^{2}+x^{2})}{cosh pi x}}\,dx=4\,log left({sqrt {2}};{frac {Gamma left({frac {3}{4}}+{frac {alpha }{2}} ight)}{Gamma left({frac {1}{4}}+{frac {alpha }{2}} ight)}} ight)}{displaystyle int _{-infty }^{infty }{frac {log(alpha ^{2}+x^{2})}{cosh pi x}}\,dx=4\,log left({sqrt {2}};{frac {Gamma left({frac {3}{4}}+{frac {alpha }{2}} ight)}{Gamma left({frac {1}{4}}+{frac {alpha }{2}} ight)}} ight)} für {displaystyle alpha geq 0}{displaystyle alpha geq 0}
ist {displaystyle {frac {pi }{8}}\,int _{-infty }^{infty }{frac {log x^{2}}{cosh pi x}}\,dx={frac {pi }{2}}\,log left({sqrt {2}}\,\,{frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}{displaystyle {frac {pi }{8}}\,int _{-infty }^{infty }{frac {log x^{2}}{cosh pi x}}\,dx={frac {pi }{2}}\,log left({sqrt {2}}\,\,{frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}. Also ist {displaystyle I={frac {pi }{2}}\,log left({sqrt {2pi }};{frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}{displaystyle I={frac {pi }{2}}\,log left({sqrt {2pi }};{frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}.
2. Beweis
In der Formel {displaystyle int _{0}^{1}{frac {log log left({frac {1}{x}}
ight)}{1+2cos alpha pi cdot x+x^{2}}}\,dx={frac {pi }{2sin alpha pi }}left(alpha log 2pi +log {frac {Gamma left({frac {1}{2}}+{frac {alpha }{2}}
ight)}{Gamma left({frac {1}{2}}-{frac {alpha }{2}}
ight)}}
ight)}{displaystyle int _{0}^{1}{frac {log log left({frac {1}{x}}
ight)}{1+2cos alpha pi cdot x+x^{2}}}\,dx={frac {pi }{2sin alpha pi }}left(alpha log 2pi +log {frac {Gamma left({frac {1}{2}}+{frac {alpha }{2}}
ight)}{Gamma left({frac {1}{2}}-{frac {alpha }{2}}
ight)}}
ight)}
setze {displaystyle alpha ={frac {1}{2}}\,:quad int _{0}^{1}{frac {log log left({frac {1}{x}} ight)}{1+x^{2}}}\,dx={frac {pi }{2}}left(log {sqrt {2pi }}+log {frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}{displaystyle alpha ={frac {1}{2}}\,:quad int _{0}^{1}{frac {log log left({frac {1}{x}} ight)}{1+x^{2}}}\,dx={frac {pi }{2}}left(log {sqrt {2pi }}+log {frac {Gamma left({frac {3}{4}} ight)}{Gamma left({frac {1}{4}} ight)}} ight)}
Durch die Substitution {displaystyle xmapsto cot x}{displaystyle xmapsto cot x} ergibt sich die besagte Gleichung.