Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,log)

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{displaystyle int _{0}^{1}{frac {log(1+x)-log 2}{1+x^{2}}}\,dx=-{frac {pi }{8}}log 2}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log(1+x)-log 2}{1-x^{2}}}\,dx=-{frac {pi ^{2}}{24}}}

ohne Beweis

 

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{displaystyle int _{1}^{infty }{frac {log(1+x)-log 2}{1+x^{2}}}\,dx=G-{frac {pi }{8}}log 2}

ohne Beweis

 

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{displaystyle int _{1}^{infty }{frac {log(1+x)-log 2}{1-x^{2}}}\,dx=-{frac {pi ^{2}}{12}}}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log x}{1+x^{2}}}\,dx=-G}

Beweis

Aus {displaystyle {frac {log x}{1+x^{2}}}=sum _{n=0}^{infty }(-1)^{n}\,x^{2n}\,log x}

folgt {displaystyle int _{0}^{1}{frac {log x}{1+x^{2}}}dx=sum _{n=0}^{infty }(-1)^{n}int _{0}^{1}x^{2n}log x\,dx=sum _{n=0}^{infty }(-1)^{n}\,{frac {-1}{(2n+1)^{2}}}=-G}.

 

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{displaystyle int _{0}^{1}{frac {log x}{1-x^{2}}}\,dx=-{frac {pi ^{2}}{8}}}

Beweis

Aus {displaystyle {frac {log x}{1-x^{2}}}=sum _{n=0}^{infty }x^{2n}\,log x}

folgt {displaystyle int _{0}^{1}{frac {log x}{1-x^{2}}}dx=sum _{n=0}^{infty }int _{0}^{1}x^{2n}log x\,dx=sum _{n=0}^{infty }{frac {-1}{(2n+1)^{2}}}=-{frac {pi ^{2}}{8}}}.

 

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{displaystyle int _{0}^{infty }{frac {log(1+x+x^{2})}{1+x^{2}}}\,dx={frac {pi }{3}}log(2+{sqrt {3}})+{frac {4}{3}}\,G}

Beweis

In der Formel {displaystyle int _{0}^{infty }{frac {log(1+2sin alpha cdot x+x^{2})}{1+x^{2}}}\,dx=pi log left(2cos {frac {alpha }{2}} ight)+alpha log left( an {frac {alpha }{2}} ight)+2sum _{k=0}^{infty }{frac {sin(2k+1)alpha }{(2k+1)^{2}}}} setze {displaystyle alpha ={frac {pi }{6}}}.

Wegen {displaystyle pi log left(2cos {frac {pi }{12}} ight)={frac {pi }{2}}left(left(2cos {frac {pi }{12}} ight)^{2} ight)={frac {pi }{2}}log(2+{sqrt {3}})\,,;{frac {pi }{6}}log left( an {frac {pi }{12}} ight)=-{frac {pi }{6}}log left(cot {frac {pi }{12}} ight)=-{frac {pi }{6}}log(2+{sqrt {3}})}

und {displaystyle 2sum _{k=0}^{3N}{frac {sin(2k+1){frac {pi }{6}}}{(2k+1)^{2}}}=sum _{k=0}^{3N}{frac {(-1)^{k}}{(2k+1)^{2}}}+3sum _{k=0}^{N-1}{frac {(-1)^{k}}{(6k+3)^{2}}} o G+{frac {G}{3}}={frac {4}{3}}\,G}

ist dann {displaystyle int _{0}^{infty }{frac {log(1+x+x^{2})}{1+x^{2}}}\,dx={frac {pi }{3}}log(2+{sqrt {3}})+{frac {4}{3}}\,G}.

 

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{displaystyle int _{0}^{1}log(-log x)\,dx=-gamma }

Beweis

Differenziere {displaystyle Gamma (z)=int _{0}^{infty }t^{z-1}\,e^{-t}\,dt}.

{displaystyle Gamma '(z)=int _{0}^{infty }t^{z-1}\,log(t)\,e^{-t}\,dt}.

und setze {displaystyle z=1\,}.

{displaystyle -gamma =Gamma '(1)=int _{0}^{infty }log(t)\,e^{-t}\,dt}.

Dies ist nach Substitution {displaystyle tmapsto -log x} gleich {displaystyle int _{0}^{1}log(-log x)\,dx}.

 

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{displaystyle int _{0}^{1}left({frac {2log x}{x^{2}-4x+8}}-{frac {3log x}{x^{2}+2x+2}} ight)dx=G}

ohne Beweis

 

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{displaystyle int _{0}^{infty }{frac {log(x+1)}{log ^{2}x+pi ^{2}}}\,{frac {dx}{x^{2}}}=gamma }

Beweis

Betrachte die Formel {displaystyle int _{0}^{infty }{frac {1}{x+1-u}}cdot {frac {1}{log ^{2}x+pi ^{2}}}\,dx={frac {1}{u}}+{frac {1}{log(1-u)}}}.

Wegen {displaystyle int _{0}^{1}{frac {1}{x+1-u}}\,du=left[-log(x+1-u) ight]_{0}^{1}=log left(1+{frac {1}{x}} ight)}

und {displaystyle int _{0}^{1}left[{frac {1}{u}}+{frac {1}{log(1-u)}} ight]du=gamma } ist {displaystyle int _{0}^{infty }{frac {log left(1+{frac {1}{x}} ight)}{log ^{2}x+pi ^{2}}}\,dx=gamma }.

Nach der Substitution {displaystyle xmapsto {frac {1}{x}}} erhält man das gesuchte Integral.

 

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{displaystyle int _{0}^{1}{frac {log left(1+x^{2+{sqrt {3}}}\, ight)}{1+x}}\,dx={frac {pi ^{2}}{12}}cdot left(1-{sqrt {3}}\, ight)+log(2)cdot log left(1+{sqrt {3}}\, ight)}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log left(1+x^{4+{sqrt {15}}}\, ight)}{1+x}}\,dx={frac {pi ^{2}}{12}}cdot left(2-{sqrt {15}}\, ight)+log left({frac {1+{sqrt {5}}}{2}} ight)cdot log left(2+{sqrt {3}}\, ight)+log(2)cdot log left({sqrt {3}}+{sqrt {5}}\, ight)}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log left(1+x^{6+{sqrt {35}}}\, ight)}{1+x}}\,dx={frac {pi ^{2}}{12}}cdot left(3-{sqrt {35}}\, ight)+log left({frac {1+{sqrt {5}}}{2}} ight)cdot log left(8+3{sqrt {7}}\, ight)+log(2)cdot log left({sqrt {5}}+{sqrt {7}}\, ight)}

ohne Beweis

 

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{displaystyle int _{0}^{1}log(1-x)\,log(x)\,log(1+x)\,dx=-6+{frac {5pi ^{2}}{12}}-(log 2)^{2}+4cdot log 2-{frac {pi ^{2}}{2}}log 2+{frac {21}{8}}zeta (3)}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log(1+x)\,log x}{1+x}}\,dx=-{frac {zeta (3)}{8}}}

Beweis

Wegen {displaystyle {frac {log(1+x)}{1+x}}=sum _{k=1}^{infty }(-1)^{k-1}\,H_{k} x^{k}} ist

{displaystyle int _{0}^{1}{frac {log(1+x)\,log x}{1+x}}\,dx=sum _{k=1}^{infty }(-1)^{k-1}\,H_{k}\,int _{0}^{1}x^{k}\,log x\,\,dx=sum _{k=1}^{infty }(-1)^{k}\,{frac {H_{k}}{(k+1)^{2}}}=-{frac {zeta (3)}{8}}}.

 

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{displaystyle int _{0}^{1}{frac {log(1+x)\,log x}{x}}\,dx=-{frac {3}{4}}\,zeta (3)}

Beweis

{displaystyle log(1+x)=sum _{k=1}^{infty }(-1)^{k-1}\,{frac {x^{k}}{k}}}

{displaystyle int _{0}^{1}{frac {log(1+x)\,log x}{x}}\,dx=sum _{k=1}^{infty }(-1)^{k-1}\,{frac {1}{k}}\,int _{0}^{1}x^{k-1}\,log x\,dx=sum _{k=1}^{infty }{frac {(-1)^{k}}{k^{3}}}=-{frac {3}{4}}\,zeta (3)}

 

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{displaystyle int _{0}^{1}{frac {log(1+x)\,log x}{1-x}}\,dx=-{frac {pi ^{2}}{4}}log 2+zeta (3)}

Beweis

{displaystyle I:=int _{0}^{1}{frac {log(1+x)\,log x}{1-x}}\,dx=sum _{n=1}^{infty }{frac {(-1)^{n-1}}{n}}int _{0}^{1}{frac {x^{n}\,log x}{1-x}}\,dx},

wobei {displaystyle int _{0}^{1}{frac {x^{n}\,log x}{1-x}}\,dx=sum _{k=0}^{infty }int _{0}^{1}x^{n+k}\,log x\,dx=sum _{k=0}^{infty }{frac {-1}{(n+k+1)^{2}}}=sum _{k=1}^{n}{frac {1}{k^{2}}}-{frac {pi ^{2}}{6}}} ist.

{displaystyle I=sum _{n=1}^{infty }{frac {(-1)^{n-1}}{n}}left(H_{n,2}-{frac {pi ^{2}}{6}} ight)=underbrace {sum _{n=1}^{infty }{frac {(-1)^{n-1}}{n}}\,H_{n,2}} _{=-{frac {pi ^{2}}{12}}log 2+zeta (3)}-{frac {pi ^{2}}{6}}log 2=-{frac {pi ^{2}}{4}}log 2+zeta (3)}

 

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{displaystyle int _{0}^{1}{frac {log(1-x)\,log x}{1+x}}\,dx=-{frac {pi ^{2}}{4}}log 2+{frac {13}{8}}\,zeta (3)}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log(1-x)\,log x}{x}}\,dx=zeta (3)}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log(1-x)\,log x}{1-x}}\,dx=zeta (3)}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log(1-x)\,log(1+x)}{x}}\,dx=-{frac {5}{8}}\,zeta (3)}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log(1-x)\,log(1+x)}{1+x}}\,dx={frac {(log 2)^{2}}{8}}-{frac {pi ^{2}}{12}}log 2+{frac {zeta (3)}{8}}}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {log ^{2n}x}{1+x^{2}}}\,dx={frac {1}{2}}\,|E_{2n}|\,left({frac {pi }{2}} ight)^{2n+1}qquad nin mathbb {Z} ^{geq 0}}

ohne Beweis

 

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{displaystyle int _{0}^{infty }{frac {log ^{n-1}x}{1+x^{2}}}\,dx=|E_{n-1}|\,left({frac {pi }{2}} ight)^{n}qquad nin mathbb {Z} ^{geq 1}}

Beweis

In der Formel {displaystyle int _{0}^{infty }{frac {x^{alpha -1}}{1+x^{eta }}}\,dx={frac {pi }{eta }}csc left({frac {alpha pi }{eta }} ight)} setze {displaystyle eta =2\,} und verschiebe {displaystyle alpha \,} um {displaystyle 1\,} nach rechts.

{displaystyle int _{0}^{infty }{frac {x^{alpha }}{1+x^{2}}}\,dx={frac {pi }{2}}\,csc left({frac {(alpha +1)pi }{2}} ight)={frac {pi }{2}}sec left({frac {alpha pi }{2}} ight)={frac {pi }{2}}sum _{k=0}^{infty }|E_{k}|\,{frac {1}{k!}}\,left({frac {alpha pi }{2}} ight)^{k}=sum _{k=0}^{infty }|E_{k}|\,left({frac {pi }{2}} ight)^{k+1}\,{frac {alpha ^{k}}{k!}}}

Differenziere {displaystyle n-1\,} mal nach {displaystyle alpha \,}

{displaystyle int _{0}^{infty }{frac {x^{alpha }\,log ^{n-1}x}{1+x^{2}}}\,dx=sum _{k=n-1}^{infty }|E_{k}|\,left({frac {pi }{2}} ight)^{k+1}\,{frac {alpha ^{k-n+1}}{(k-n+1)!}}}

Und setze {displaystyle alpha =0\,}

{displaystyle int _{0}^{infty }{frac {log ^{n-1}x}{1+x^{2}}}\,dx=|E_{n-1}|\,left({frac {pi }{2}} ight)^{n}}

 

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{displaystyle int _{0}^{infty }{frac {log ^{n-1}x}{1-x^{2}}}\,dx={frac {2^{n}(1-2^{n})|B_{n}|}{n}}\,left({frac {pi }{2}} ight)^{n}qquad nin mathbb {Z} ^{geq 1}}

ohne Beweis

 

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{displaystyle int _{0}^{infty }{frac {dx}{(x+1)^{n}\,(log ^{2}x+pi ^{2})}}=C_{n}} Fontana-Zahlen genügen der Rekursion: {displaystyle quad C_{0}=-1,quad sum _{k=0}^{n-1}{frac {C_{k}}{n-k}}=0}

ohne Beweis

 

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{displaystyle int _{0}^{infty }{frac {1}{x+1-u}}cdot {frac {1}{log ^{2}x+pi ^{2}}}\,dx={frac {1}{u}}+{frac {1}{log(1-u)}}qquad uin mathbb {C} ^{ imes }setminus mathbb {R} ^{geq 1}}

Beweis

Die Funktion {displaystyle f(z)={frac {1}{z+1-u}}cdot {frac {1}{log(-z)}}} ist auf {displaystyle mathbb {C} setminus mathbb {R} ^{geq 0}} meromorph.

Die Polstellen liegen bei {displaystyle -1\,} und {displaystyle u-1\,}. Dabei ist {displaystyle { ext{res}}(f,-1)={frac {1}{u}}} und {displaystyle { ext{res}}(f,u-1)={frac {1}{log(1-u)}}}.

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=2pi i\,left({frac {1}{u}}+{frac {1}{log(1-u)}} ight)}.

Aus {displaystyle L(C_{R})sim 2pi R\,} und {displaystyle M(C_{R})=max _{zin C_{R}}|f(z)|sim {frac {1}{R\,log R}}} folgt {displaystyle left|int _{C_{R}}f\,dz ight|leq L(C_{R})\,M(C_{R})sim {frac {2pi }{log R}}}.

Daher geht {displaystyle int _{C_{R}}fdz} gegen null für {displaystyle R o infty \,}.

Für {displaystyle z\,}, nahe bei 0, ist {displaystyle log(-z)\,} groß, und somit {displaystyle f(z)={frac {1}{z+1-u}}cdot {frac {1}{log(-z)}}} klein.

Daher geht {displaystyle int _{c_{varepsilon }}f\,dz} für {displaystyle varepsilon o 0+\,} auch gegen null.

Im Grenzübergang {displaystyle R o infty ,varepsilon o 0+\,} ergibt sich:

{displaystyle int _{0}^{infty }{frac {1}{x+1-u}}left({frac {1}{log(-x-i0^{+})}}-{frac {1}{log(-x+i0^{+})}} ight)dx=2pi ileft({frac {1}{u}}+{frac {1}{log(1-u)}} ight)}.

Dabei ist {displaystyle {frac {1}{log x-ipi }}-{frac {1}{log x+ipi }}={frac {2pi i}{log ^{2}x+pi ^{2}}}}, und somit gilt

{displaystyle int _{0}^{infty }{frac {1}{x+1-u}}cdot {frac {1}{log ^{2}x+pi ^{2}}}\,dx={frac {1}{u}}+{frac {1}{log(1-u)}}}.

 

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{displaystyle int _{0}^{1}{frac {left(log \,{frac {1}{x}} ight)^{z-1}}{1+x}}\,dx=eta (z)\,Gamma (z)qquad { ext{Re}}(z)>0}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {left(log \,{frac {1}{x}} ight)^{z-1}}{1-x}}\,dx=zeta (z)\,Gamma (z)qquad { ext{Re}}(z)>1}

ohne Beweis

 

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{displaystyle int _{0}^{1}{frac {x^{alpha -1}-x^{-alpha }}{(x+1)\,log x}}\,dx=log left( an {frac {alpha pi }{2}} ight)qquad 0<mathrm {Re} (alpha )<1}

Beweis

{displaystyle int _{0}^{1}{frac {x^{-alpha }}{1+x}}\,dx} ist nach Substitution {displaystyle xmapsto {frac {1}{x}}} gleich {displaystyle int _{1}^{infty }{frac {x^{alpha -1}}{1+x}}\,dx}.

Also ist {displaystyle int _{0}^{1}{frac {x^{alpha -1}+x^{-alpha }}{1+x}}\,dx=int _{0}^{1}{frac {x^{alpha -1}}{1+x}}\,dx+int _{1}^{infty }{frac {x^{alpha -1}}{1+x}}\,dx}   {displaystyle =int _{0}^{infty }{frac {x^{alpha -1}}{1+x}}\,dx={frac {pi }{sin alpha pi }}}.

Integriere nach {displaystyle alpha \,}:

{displaystyle int _{0}^{1}{frac {x^{alpha -1}-x^{-alpha }}{(1+x)log x}}\,dx=log left( an {frac {alpha pi }{2}} ight)+C}

Dass dabei {displaystyle C=0\,} ist, erkennt man, wenn man {displaystyle alpha ={frac {1}{2}}} setzt.

 

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{displaystyle int _{0}^{1}left(log {frac {1}{x}} ight)^{z-1}dx=Gamma (z)qquad { ext{Re}}(z)>0}

Beweis

In der Formel {displaystyle Gamma (z)=int _{0}^{infty }t^{z-1}e^{-t}\,dt} substituiere {displaystyle t=-log x\,}:

{displaystyle Gamma (z)=int _{0}^{1}left(-log x ight)^{z-1}dx}

 

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{displaystyle int _{0}^{1}left({frac {log x}{a+1-x}}-{frac {log x}{a+x}} ight)dx={frac {1}{2}}left(log a-log(a+1) ight)^{2}qquad forall ain mathbb {C} setminus [-1,0]}

Beweis

Definiert man {displaystyle F:\,]0,1] o mathbb {C} } als {displaystyle F(x)={frac {xlog x}{a+x}}-log(a+x)},

so ist {displaystyle F'(x)={frac {log x+1}{a+x}}-{frac {xlog x}{(a+x)^{2}}}-{frac {1}{a+x}}}

{displaystyle ={frac {alog x+a+xlog x+x-xlog x-(a+x)}{(a+x)^{2}}}={frac {alog x}{(a+x)^{2}}}}.

Also ist {displaystyle aint _{0}^{1}{frac {log x}{(a+x)^{2}}}\,dx=F(1)-F(0+)=log a-log(a+1)}.


Definiert man {displaystyle G:\,]0,1] o mathbb {C} } als {displaystyle G(x)={frac {xlog x}{a+1-x}}+log(a+1-x)},

so ist {displaystyle G'(x)={frac {log x+1}{a+1-x}}+{frac {xlog x}{(a+1-x)^{2}}}-{frac {1}{a+1-x}}}

{displaystyle ={frac {(a+1)log x+a+1-xlog x-x+xlog x-(a+1-x)}{(a+1-x)^{2}}}={frac {(a+1)log x}{(a+1-x)^{2}}}}

Also ist {displaystyle (a+1)int _{0}^{1}{frac {log x}{(a+1-x)^{2}}}\,dx=G(1)-G(0+)=log a-log(a+1)}.


Mit den beiden Integralen erhält man {displaystyle int _{0}^{1}left({frac {log x}{(a+x)^{2}}}-{frac {log x}{(a+1-x)^{2}}} ight)dx=left(log a-log(a+1) ight)left({frac {1}{a}}-{frac {1}{a+1}} ight)}.

Integriert man beide Seiten nach {displaystyle a\,}, so ist {displaystyle int _{0}^{1}left(-{frac {log x}{a+x}}+{frac {log x}{a+1-x}} ight)dx={frac {1}{2}}left(log a-log(a+1) ight)^{2}+C}.

Dass die Konstante {displaystyle C=0\,} sein muss, erkennt man wenn man {displaystyle a o infty \,} gehen lässt.

 

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{displaystyle int _{0}^{1}{frac {2log x}{(api )^{2}+log ^{2}x}}\,{frac {x}{1-x^{2}}}\,dx={frac {1}{2a}}+psi (a)-log aqquad { ext{Re}}(a)>0}

Beweis

{displaystyle int _{0}^{1}{frac {2log x}{(api )^{2}+log ^{2}x}}\,{frac {x}{1-x^{2}}}\,dx=int _{0}^{1}}   {displaystyle int _{0}^{infty }2sin(tlog x)\,e^{-api t}\,dt\,}   {displaystyle \,{frac {x}{1-x^{2}}}\,dx}

{displaystyle =int _{0}^{infty }int _{0}^{1}2sin(tlog x)\,{frac {x}{1-x^{2}}}\,dx\,e^{-api t}\,dtqquad { ext{//subst:}}\,\,x=e^{-u}}

{displaystyle =int _{0}^{infty }int _{0}^{infty }{frac {2sin(tu)}{1-e^{2u}}}\,du\,e^{-api t}\,dt=int _{0}^{infty }}   {displaystyle left({frac {1}{t}}-{frac {pi }{2}}coth {frac {pi t}{2}} ight)}   {displaystyle e^{-api t}\,dtqquad { ext{//subst:}}\,\,omega =pi t}

{displaystyle =int _{0}^{infty }left({frac {1}{omega }}-{frac {1}{e^{omega }-1}}-{frac {1}{2}} ight)e^{-aomega }\,domega =}   {displaystyle psi (a+1)-log a-{frac {1}{2a}}}

 

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{displaystyle int _{0}^{infty }{frac {log(1+2sin alpha \,\,x+x^{2})}{1+x^{2}}}\,dx=pi log left(2cos {frac {alpha }{2}} ight)+alpha log left( an {frac {alpha }{2}} ight)+2sum _{k=0}^{infty }{frac {sin(2k+1)alpha }{(2k+1)^{2}}}qquad 0<alpha <pi }

Beweis

Setzt man {displaystyle F(z):=int _{0}^{infty }{frac {log(1+2sin z;\,x+x^{2})}{1+x^{2}}}\,dx},

so ist {displaystyle F(0)=int _{0}^{infty }{frac {log(1+x^{2})}{1+x^{2}}}\,dx=int _{0}^{frac {pi }{2}}log(1+ an ^{2}x)\,dx=-2int _{0}^{frac {pi }{2}}log(cos x)\,dx=pi log 2}

und {displaystyle F'(z)=int _{0}^{infty }{frac {1}{1+x^{2}}}\,{frac {2xcos z}{1+2sin z;\,x+x^{2}}}\,dx={frac {cos z}{sin z}}int _{0}^{infty }{frac {1}{1+x^{2}}}\,{frac {2xsin z}{1+2sin z;\,x+x^{2}}}\,dx}

{displaystyle ={frac {cos z}{sin z}}int _{0}^{infty }left({frac {1}{1+x^{2}}}-{frac {1}{1+2sin z;\,x+x^{2}}} ight)dx={frac {pi }{2}}cdot {frac {cos z}{sin z}}-{frac {cos z}{sin z}}left[{frac {1}{cos z}}\,arctan {frac {x+sin z}{cos z}} ight]_{0}^{infty }}

{displaystyle ={frac {pi }{2}}cdot {frac {cos z}{sin z}}-{frac {pi }{2}}cdot {frac {1}{sin z}}+{frac {z}{sin z}}=-{frac {pi }{2}} an {frac {z}{2}}+{frac {z}{sin z}}}.

Nun ist {displaystyle F(alpha )-pi log 2=int _{0}^{alpha }F'(z)\,dz=pi log left(cos {frac {alpha }{2}} ight)+int _{0}^{alpha }{frac {z}{sin z}}\,dz}

und somit ist {displaystyle F(alpha )-pi log left(2cos {frac {alpha }{2}} ight)=int _{0}^{alpha }{frac {z}{sin z}}\,dz=left[zlog left( an {frac {z}{2}} ight) ight]_{0}^{alpha }-int _{0}^{alpha }log left( an {frac {z}{2}} ight)dz}

{displaystyle =alpha log left( an {frac {alpha }{2}} ight)+2int _{0}^{alpha }sum _{k=0}^{infty }{frac {cos(2k+1)z}{2k+1}}\,dz}.

Daraus folgt {displaystyle F(alpha )=pi log left(2cos {frac {alpha }{2}} ight)+alpha log left( an {frac {alpha }{2}} ight)+2sum _{k=0}^{infty }{frac {sin(2k+1)alpha }{(2k+1)^{2}}}}.

 

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{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)qquad 0<alpha <1}

Beweis

{displaystyle {frac {sin alpha pi }{1+2cos alpha pi cdot x+x^{2}}}=sum _{n=1}^{infty }(-x)^{n-1}\,sin npi alpha qquad quad |x|<1}

{displaystyle int _{0}^{1}{frac {sin alpha pi }{1+2cos alpha pi cdot x+x^{2}}}\,log log left({frac {1}{x}} ight)dx=sum _{n=1}^{infty }(-1)^{n-1}int _{0}^{1}x^{n-1}\,log log left({frac {1}{x}} ight)dx\,;sin npi alpha }

{displaystyle =sum _{n=1}^{infty }(-1)^{n}\,{frac {gamma +log n}{n}}\,sin npi alpha =gamma cdot sum _{n=1}^{infty }{frac {(-1)^{n}}{n}}\,sin npi alpha +sum _{n=1}^{infty }(-1)^{n}\,{frac {log n}{n}}\,sin npi alpha }

{displaystyle -{frac {alpha pi }{2}}\,gamma +{frac {alpha pi }{2}}left(gamma +log 2pi ight)+{frac {pi }{2}}log {frac {Gamma left({frac {1}{2}}+{frac {alpha }{2}} ight)}{Gamma left({frac {1}{2}}-{frac {alpha }{2}} ight)}}={frac {pi }{2}}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)}

 

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{displaystyle int _{0}^{1}{frac {log(1-x)}{x}}\,{frac {2z}{log ^{2}x+(2pi z)^{2}}}\,dx=-log left({frac {z!\,e^{z}}{z^{z}\,{sqrt {2pi z}}}} ight)qquad { ext{Re}}(z)>0}

Beweis

 

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{displaystyle int _{a}^{b}{frac {log x}{(x+a)(x+b)}}\,dx={frac {log(ab)}{2\,(b-a)}}\,log left({frac {(a+b)^{2}}{4ab}} ight)}

Beweis

Nach der Substitution {displaystyle xmapsto {frac {ab}{x}}} wird das Integral zu {displaystyle I=int _{a}^{b}{frac {log(ab)-log x}{(b+x)(a+x)}}\,dx}

Also ist {displaystyle 2I=int _{a}^{b}{frac {log(ab)}{(x+a)(x+b)}}\,dx={frac {log(ab)}{b-a}}int _{a}^{b}left({frac {1}{x+a}}-{frac {1}{x+b}} ight)\,dx}

{displaystyle ={frac {log(ab)}{b-a}}\,{Big [}log(x+a)-log(x+b){Big ]}_{a}^{b}={frac {log(ab)}{b-a}}log left({frac {(a+b)^{2}}{4ab}} ight)}.

 

2.2Bearbeiten

{displaystyle int _{0}^{infty }{frac {log x}{(x+a)(x+b)}}\,dx={frac {log ^{2}(a)-log ^{2}(b)}{2\,(a-b)}}}

Beweis

Nach der Substitution {displaystyle xmapsto {frac {ab}{x}}} wird das Integral zu {displaystyle I=int _{0}^{infty }{frac {log(ab)-log x}{(b+x)(a+x)}}\,dx}.

Also ist {displaystyle 2I=int _{0}^{infty }{frac {log(ab)}{(x+a)(x+b)}}\,dx={frac {log(ab)}{a-b}}int _{0}^{infty }left({frac {1}{x+b}}-{frac {1}{x+a}} ight)\,dx}

{displaystyle ={frac {log(a)+log(b)}{a-b}}\,underbrace {{Big [}log(x+b)-log(x+a){Big ]}_{0}^{infty }} _{log(a)-log(b)}={frac {log ^{2}(a)-log ^{2}(b)}{a-b}}}.