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

1.1Bearbeiten

{displaystyle {frac {1}{2pi i}}int _{n-iinfty }^{n+iinfty }{frac {pi }{x\,cos pi x}}\,dx=2sum _{k=n}^{infty }{frac {(-1)^{k}}{2k+1}}qquad nin mathbb {Z} ^{>0}}

ohne Beweis

 

1.2Bearbeiten

{displaystyle int _{-infty }^{infty }{frac {cos alpha x}{1+x^{2}}}\,dx=pi ;e^{-alpha }qquad alpha geq 0}

1. Beweis

Es sei {displaystyle R>1,alpha geq 0,f(z)={frac {e^{ialpha z}}{1+z^{2}}},}

{displaystyle K_{R}\,} der Halbkreis von {displaystyle R\,} nach {displaystyle -R\,}

und {displaystyle gamma _{R}=[-R,R]+K_{R}\,}

der geschlossene halbmondförmige Integrationsweg.

Für alle {displaystyle z=x+iyin gamma _{R}} ist der Imaginärteil {displaystyle ygeq 0}

und somit {displaystyle left|e^{ialpha z} ight|leq e^{-alpha y}leq 1}.

Nun gilt {displaystyle left|int _{K_{R}}f\,dz ight|leq |K_{R}|cdot max _{zin K_{R}}|f(z)|leq Rpi \,{frac {1}{R^{2}-1}} o 0} für {displaystyle R o infty \,}.

Also ist {displaystyle int _{-infty }^{infty }f\,dz=lim _{R o infty }oint _{gamma _{R}}f\,dz=2pi i\,{ ext{res}}(f,i)=pi \,e^{-alpha }} und somit {displaystyle int _{-infty }^{infty }{frac {cos alpha x}{1+x^{2}}}\,dx=pi \,e^{-alpha }}.

2. Beweis

Aus {displaystyle {frac {x}{1+x^{2}}}=int _{0}^{infty }sin tx\,e^{-t}\,dt} folgt {displaystyle 2\,{frac {cos alpha x}{1+x^{2}}}=int _{0}^{infty }{frac {2\,sin tx\,cos alpha x}{x}}\,e^{-t}\,dt}.

Und das ist {displaystyle int _{0}^{infty }{frac {sin(t+alpha )x+sin(t-alpha )x}{x}}\,e^{-t}\,dt}.

Also ist {displaystyle 2int _{0}^{infty }{frac {cos alpha x}{1+x^{2}}}\,dx=pi int _{0}^{infty }left[{ ext{sgn}}(t+alpha )+{ ext{sgn}}(t-alpha ) ight]\,e^{-t}\,dt=pi int _{alpha }^{infty }e^{-t}\,dt=pi \,e^{-alpha }}.

 

1.3Bearbeiten

{displaystyle int _{-infty }^{infty }{frac {cos alpha x}{1-x^{2}}}\,dx=pi ;sin alpha qquad alpha geq 0}

ohne Beweis

 

1.4Bearbeiten

{displaystyle int _{-infty }^{infty }{frac {|cos alpha x|}{1+x^{2}}}\,dx=4cosh alpha \,;arctan e^{-alpha }qquad alpha >0}

Beweis

Aus der Fourierreihe {displaystyle |cos alpha x|={frac {2}{pi }}+{frac {2}{pi }}sum _{n=1}^{infty }(-1)^{n}left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)cos 2nalpha x} ergibt sich

{displaystyle int _{-infty }^{infty }{frac {|cos alpha x|}{1+x^{2}}}\,dx=2+2sum _{n=1}^{infty }(-1)^{n}left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)\,underbrace {{frac {1}{pi }}int _{-infty }^{infty }{frac {cos 2nalpha x}{1+x^{2}}}\,dx} _{e^{-2nalpha }}}

{displaystyle =2+2sum _{n=1}^{infty }(-1)^{n}\,{frac {(e^{-alpha })^{2n}}{2n+1}}-2sum _{n=1}^{infty }(-1)^{n}\,{frac {(e^{-alpha })^{2n}}{2n-1}}=2sum _{n=0}^{infty }(-1)^{n}\,{frac {(e^{-alpha })^{2n}}{2n+1}}+2sum _{n=0}^{infty }(-1)^{n}\,{frac {(e^{-alpha })^{2n+2}}{2n+1}}}

{displaystyle =2\,(e^{alpha }+e^{-alpha })sum _{n=0}^{infty }(-1)^{n}\,{frac {(e^{-alpha })^{2n+1}}{2n+1}}=4cosh alpha \,\,{ ext{arctan}}\,e^{-alpha }}.

 

1.5Bearbeiten

{displaystyle int _{-infty }^{infty }cos left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)dx={sqrt {frac {pi }{2}}}\,e^{-2alpha }}

Beweis

Die Funktion {displaystyle f(z)=exp left(-z^{2}-{frac {alpha ^{2}}{z^{2}}} ight)} ist auf dem Kreissektor {displaystyle S_{R}:=left{re^{ivarphi }\,{Big |}\,0<varphi <{frac {pi }{4}};,;0<r<R ight}} holomorph. 
Auf dem Kreisbogen {displaystyle K_{R}\,} fällt {displaystyle f\,} für {displaystyle R o infty \,} exponentiell gegen null ab.

Daher ist {displaystyle lim _{R o infty }int _{K_{R}}fdz=0} und somit {displaystyle J:=int _{0}^{infty }f(x)dx=int _{0}^{infty }fleft(e^{i{frac {pi }{4}}}x ight)\,e^{i{frac {pi }{4}}}\,dx}.

Also ist {displaystyle J:=int _{0}^{infty }e^{-ileft(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)}\,{frac {1+i}{sqrt {2}}}\,dx=int _{0}^{infty }left[cos left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)-isin left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight) ight]\,{frac {1+i}{sqrt {2}}}\,dx}

{displaystyle ={frac {1}{sqrt {2}}}int _{0}^{infty }left[cos left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)+sin left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight) ight]dx+{frac {i}{sqrt {2}}}int _{0}^{infty }left[cos left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)-sin left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight) ight]dx}.

Nun ist {displaystyle {sqrt {2}}int _{0}^{infty }cos left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)dx={ ext{Re}}(J)+{ ext{Im}}(J)={frac {sqrt {pi }}{2}}\,e^{-2alpha }+0}

und {displaystyle {sqrt {2}}int _{0}^{infty }sin left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)dx={ ext{Re}}(J)-{ ext{Im}}(J)={frac {sqrt {pi }}{2}}\,e^{-2alpha }-0}.

Also sind {displaystyle int _{-infty }^{infty }cos left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)dx} und {displaystyle int _{-infty }^{infty }sin left(x^{2}-{frac {alpha ^{2}}{x^{2}}} ight)dx} jeweils {displaystyle {sqrt {frac {pi }{2}}}\,e^{-2alpha }}.

 

1.6Bearbeiten

{displaystyle int _{-infty }^{infty }cos left(x^{2}+{frac {alpha ^{2}}{x^{2}}} ight)dx={sqrt {frac {pi }{2}}}\,(cos 2alpha -sin 2alpha )}

Beweis

{displaystyle int _{-infty }^{infty }cos left(x^{2}+{frac {alpha ^{2}}{x^{2}}} ight)dx=int _{-infty }^{infty }cos left(left(x-{frac {alpha }{x}} ight)^{2}+2alpha ight)dx}

ist nach der Formel {displaystyle int _{-infty }^{infty }fleft(x-{frac {b}{x}} ight)dx=int _{-infty }^{infty }f(x)dx}, gleich

{displaystyle int _{-infty }^{infty }cos(x^{2}+2alpha )dx=int _{-infty }^{infty }left(cos x^{2}\,cos 2alpha -sin x^{2}\,sin 2alpha ight)dx={sqrt {frac {pi }{2}}}\,cos 2alpha -{sqrt {frac {pi }{2}}}\,sin 2alpha }.

 

1.7Bearbeiten

{displaystyle J_{ u }(z)={frac {2}{Gamma left({frac {1}{2}} ight)Gamma left( u +{frac {1}{2}} ight)}}left({frac {z}{2}} ight)^{ u }int _{0}^{1}cos(zx)\,(1-x^{2})^{ u -{frac {1}{2}}}\,dxqquad { ext{Re}}( u )>-{frac {1}{2}}}

Beweis (Poissonsche Darstellung der Besselfunktion)

Die Formel ist äquivalent zur Formel

{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}}}

 

1.8Bearbeiten

{displaystyle int _{0}^{frac {pi }{2}}cos ^{2n}(x)\,dx={frac {1}{2^{2n}}}\,{2n choose n}\,{frac {pi }{2}}qquad nin mathbb {N} }

Beweis

Für {displaystyle m>n} ist {displaystyle sum _{k=0}^{m-1}left(2cos {frac {kpi }{m}} ight)^{2n}=sum _{k=0}^{m-1}left[exp left(ipi cdot {frac {k}{m}} ight)+exp left(-ipi cdot {frac {k}{m}} ight) ight]^{2n}}

{displaystyle =sum _{k=0}^{m-1}sum _{ell =0}^{2n}{2n choose ell }\,exp left(ipi cdot {frac {k}{m}}cdot ell ight)exp left(-ipi cdot {frac {k}{m}}cdot (2n-ell ) ight)=sum _{ell =0}^{2n}{2n choose ell }sum _{k=0}^{m-1}exp left(2pi icdot {frac {ell -n}{m}}cdot k ight)}.

Nun ist {displaystyle sum _{k=0}^{m-1}exp left(2pi icdot {frac {ell -n}{m}}cdot k ight)=left{{egin{matrix}0&{ ext{für}}&ell eq n\m&{ ext{für}}&ell =nend{matrix}} ight.}.

Also ist {displaystyle sum _{k=0}^{m-1}left(2cos {frac {kpi }{m}} ight)^{2n}={2n choose n}cdot m} und somit ist {displaystyle {frac {1}{m}}sum _{k=0}^{m-1}cos ^{2n}left({frac {kpi }{m}} ight)={frac {1}{2^{2n}}}{2n choose n}}.

Durch den Grenzübergang {displaystyle m o infty } erhält man {displaystyle int _{0}^{1}cos ^{2n}(pi x)\,dx={frac {1}{2^{2n}}}{2n choose n}}.

Nach Substitution {displaystyle xmapsto {frac {x}{pi }}} ist {displaystyle int _{0}^{pi }cos ^{2n}(x)\,dx={frac {1}{2^{2n}}}{2n choose n}cdot pi }.

Nachdem {displaystyle int _{0}^{frac {pi }{2}}cos ^{2n}(x)\,dx=int _{frac {pi }{2}}^{pi }cos ^{2n}(x)\,dx} ist, folgt daraus die Behauptung.

 

1.9Bearbeiten

{displaystyle int _{0}^{frac {pi }{2}}cos ^{2n+1}(x)\,dx={frac {1}{2n+1}}left[{frac {1}{2^{2n}}}\,{2n choose n} ight]^{-1}qquad nin mathbb {N} }

ohne Beweis

 

2.1Bearbeiten

{displaystyle int _{0}^{pi }cos nx\,cos mx\,dx=delta _{mn}{frac {pi }{2}}qquad n,min mathbb {Z} ^{geq 1}}

Beweis

Aus der Formel {displaystyle 2\,cos nx\,cos mx=cos(n-m)x+cos(n+m)x} folgt

{displaystyle 2int _{0}^{pi }cos nx\,cos mx\,dx=underbrace {int _{0}^{pi }cos(n-m)x\,dx} _{=delta _{nm}\,pi }+underbrace {int _{0}^{pi }cos(n+m)x\,dx} _{=0}}.

 

2.2Bearbeiten

{displaystyle {frac {1}{pi }}int _{0}^{pi }x^{2}\,cos nx\,cos mx\,dx=left{{egin{matrix}{frac {pi ^{2}}{6}}pm {frac {1}{4nm}}&,&n=m\\{frac {(-1)^{n-m}}{(n-m)^{2}}}pm {frac {(-1)^{n+m}}{(n+m)^{2}}}&,&n eq mend{matrix}} ight.qquad n,min mathbb {Z} ^{>0}}

ohne Beweis

 

2.3Bearbeiten

{displaystyle int _{0}^{pi }{frac {cos nx-cos na}{cos x-cos a}}\,dx=pi \,{frac {sin na}{sin a}}qquad nin mathbb {N} \,,\,ain mathbb {C} }

Beweis

Die Funktion {displaystyle f_{n}(x)={frac {cos nx-cos na}{cos x-cos a}}}

erfüllt die Rekursion {displaystyle f_{n+1}(x)-2cos acdot f_{n}(x)+f_{n-1}(x)=2cos nx}.

Begründung:

{displaystyle cos(n+1)x-cos(n+1)a\,+\,cos(n-1)x-cos(n-1)a}

{displaystyle =cos(n+1)x+cos(n-1)x\,-\,cos(n+1)a-cos(n-1)a}

{displaystyle =2cos nxcdot cos x-2cos nacdot cos a=2cos nxcdot (cos x-cos a)+2cos acdot (cos nx-cos na)}

{displaystyle Rightarrow {frac {cos(n+1)x-cos(n+1)a}{cos x-cos a}}+{frac {cos(n-1)x-cos(n-1)a}{cos x-cos a}}=2cos nx+2cos acdot {frac {cos nx-cos na}{cos x-cos a}}}



Also ist {displaystyle s_{n+1}-2cos acdot s_{n}+s_{n-1}=0}, wobei {displaystyle s_{n}=int _{0}^{pi }f_{n}(x)\,dx} sein soll.

Das Polynom {displaystyle x^{2}-2cos acdot x+1=0} besitzt die Wurzeln {displaystyle x_{1/2}=e^{pm ia}\,}.

Daher hat die Folge {displaystyle s_{n}\,} die Form {displaystyle s_{n}=C_{1}\,e^{ian}+C_{2}\,e^{-ian}=D_{1}cos na+D_{2}sin na}.

Aus {displaystyle s_{0}=0\,Rightarrow \,D_{1}=0} und {displaystyle s_{1}=pi \,Rightarrow \,D_{2}={frac {pi }{sin a}}} folgt schließlich {displaystyle s_{n}=pi \,{frac {sin na}{sin a}}}.

 

2.4Bearbeiten

{displaystyle int _{-pi }^{pi }{frac {cos nx}{1-2rcos x+r^{2}}}\,dx={frac {2pi \,r^{n}}{1-r^{2}}}qquad |r|<1\,,\,nin mathbb {Z} ^{geq 0}}

Beweis

Betrachte die Poissonsche Integralformel

{displaystyle {frac {1}{2pi }}int _{-pi }^{pi }P_{R}(r,phi -varphi )\,f(Re^{ivarphi })\,dvarphi =f(re^{iphi })}, wobei der Kern {displaystyle P_{R}(r,phi )={frac {R^{2}-r^{2}}{R^{2}-2Rrcos phi +r^{2}}}} ist.

Setzt man {displaystyle R=1\,,\,phi =0} und {displaystyle f(z)=z^{n}\,}, so ist {displaystyle P_{R}(r,phi -x)={frac {1-r^{2}}{1-2rcos x+r^{2}}}} und {displaystyle f(Re^{ix})=e^{inx}\,}.

Also ist {displaystyle int _{-pi }^{pi }{frac {cos nx+isin nx}{1-2rcos x+r^{2}}}\,dx={frac {2pi \,r^{n}}{1-r^{2}}}}. Der ungerade Anteil {displaystyle int _{-pi }^{pi }{frac {sin nx}{1-2rcos x+r^{2}}}\,dx} verschwindet dabei aus symmetriegründen.

 

2.5Bearbeiten

{displaystyle int _{0}^{infty }{frac {|cos alpha x|-|cos eta x|}{x}}\,dx=left(1-{frac {2}{pi }} ight)log {frac {eta }{alpha }}qquad alpha ,eta >0}

Beweis

Aus der Fourierreihe {displaystyle |cos alpha x|={frac {2}{pi }}+{frac {2}{pi }}sum _{n=1}^{infty }(-1)^{n}left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)cos 2nalpha x} ergibt sich

{displaystyle |cos alpha x|-|cos eta x|={frac {2}{pi }}sum _{n=1}^{infty }(-1)^{n}left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight){Big (}cos 2nalpha x-cos 2neta x{Big )}}

Also ist {displaystyle int _{0}^{infty }{frac {|cos alpha x|-|cos eta x|}{x}}\,dx={frac {2}{pi }}sum _{n=1}^{infty }(-1)^{n}left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)int _{0}^{infty }{frac {cos 2nalpha x-cos 2neta x}{x}}\,dx},

wobei das Frullanische Integral {displaystyle int _{0}^{infty }{frac {cos 2nalpha x-cos 2neta x}{x}}\,dx=log {frac {eta }{alpha }}} nicht von {displaystyle n\,} abhängt.

Und die Reihe {displaystyle sum _{n=1}^{infty }(-1)^{n}left({frac {1}{2n+1}}-{frac {1}{2n-1}} ight)} konvergiert gegen {displaystyle {frac {pi }{2}}-1}.

 

2.6Bearbeiten

{displaystyle int _{0}^{infty }{frac {|cos alpha x|-|cos eta x|}{x^{2}}}\,dx=eta -alpha qquad alpha ,eta >0}

ohne Beweis

 

2.7Bearbeiten

{displaystyle int _{-infty }^{infty }{frac {cos(alpha x)}{1+2cos heta \,x+x^{2}}}\,dx={frac {pi }{sin heta }}\,{frac {cos(alpha cos heta )}{e^{alpha sin heta }}}qquad alpha geq 0\,,\, heta in mathbb {C} setminus pi mathbb {Z} }

ohne Beweis

 

2.8Bearbeiten

{displaystyle int _{-infty }^{infty }{frac {cos(alpha x)}{x^{4}+eta ^{4}}}\,dx={frac {pi }{eta ^{3}\,{sqrt {2}}}}\,left(cos {frac {alpha eta }{sqrt {2}}}+sin {frac {alpha eta }{sqrt {2}}} ight)\,e^{-{frac {alpha eta }{sqrt {2}}}}qquad alpha ,eta >0}

ohne Beweis

 

2.9Bearbeiten

{displaystyle int _{-infty }^{infty }{frac {cos alpha x}{prod limits _{k=0}^{infty }left(1+{frac {x^{2}}{(eta +k)^{2}}} ight)}}\,dx={sqrt {pi }}\,{frac {Gamma left(eta +{frac {1}{2}} ight)}{Gamma (eta )}}\,{ ext{sech}}^{2eta }left({frac {alpha }{2}} ight)qquad alpha ,eta >0}

Beweis

Multipliziert man die Formel

{displaystyle int _{-infty }^{infty }{frac {cos alpha x}{(eta ^{2}+x^{2})((eta +1)^{2}+x^{2})cdots ((eta +n)^{2}+x^{2})}}\,dx=2pi sum _{k=0}^{n}(-1)^{k}\,{frac {(2eta -1+k)!}{(2eta +n+k)!}}\,{frac {1}{k!\,(n-k)!}}\,e^{-alpha (eta +k)}}

mit {displaystyle eta ^{2}\,(eta +1)^{2}cdots (eta +n)^{2}={frac {(eta +n)!^{2}}{Gamma ^{2}(eta )}}} durch, so ist

{displaystyle I_{n}:=int _{-infty }^{infty }{frac {cos alpha x}{prod limits _{k=0}^{n}left(1+{frac {x^{2}}{(eta +k)^{2}}} ight)}}\,dx=2pi \,{frac {Gamma (2eta )}{Gamma ^{2}(eta )}}\,sum _{k=0}^{n}(-1)^{k}\,{frac {(2eta -1+k)!}{Gamma (2eta )\,k!}}\,{frac {(eta +n)!^{2}}{(2eta +n+k)!\,(n-k)!}}\,e^{-alpha (eta +k)}},

wobei {displaystyle (-1)^{k}\,{frac {(2eta -1+k)!}{Gamma (2eta )\,k!}}={-2eta choose k}} ist.

Setzt man {displaystyle A_{n,k}:={frac {(eta +n)!^{2}}{(2eta +n+k)!\,(n-k)!}}} so ist {displaystyle I_{n}=2pi \,{frac {Gamma (2eta )}{Gamma ^{2}(eta )}}\,sum _{k=0}^{n}{-2eta choose k}\,A_{n,k}\,e^{-alpha k}\,e^{-alpha eta }}.

Mit einem {displaystyle 0<M<n\,} lässt sich letzte Summe folgendermaßen aufspalten:

{displaystyle 2pi \,{frac {Gamma (2eta )}{Gamma ^{2}(eta )}}left(sum _{k=0}^{M-1}{-2eta choose k}\,A_{n,k}\,e^{-alpha k}+sum _{k=M}^{n}{-2eta choose k}\,A_{n,k}\,e^{-alpha k} ight)\,e^{-alpha eta }}

Die Folge {displaystyle (A_{n,0}\,,\,A_{n,1}\,,\,...\,,\,A_{n,n})subset [0,1]} fällt monoton und für alle {displaystyle 0leq k<M} gilt {displaystyle lim _{n o infty }A_{n,k}=1\,}.

Also ist {displaystyle I:=lim _{n o infty }I_{n}=2pi {frac {Gamma (2eta )}{Gamma ^{2}(eta )}}\,left(sum _{k=0}^{M-1}{-2eta choose k}\,e^{-alpha k}+varepsilon _{M} ight)\,e^{-alpha eta }}

Für {displaystyle M o infty \,} geht {displaystyle varepsilon _{M}\,} gegen null und {displaystyle sum _{k=0}^{M-1}{-2eta choose k}\,e^{-alpha k}} konvergiert gegen die Binomialreihenentwicklung von {displaystyle (1+e^{-alpha })^{-2eta }\,}.

Also ist {displaystyle I=2pi \,{frac {Gamma (2eta )}{Gamma ^{2}(eta )}}\,left(1+e^{-alpha } ight)^{-2eta }\,left(e^{frac {alpha }{2}} ight)^{-2eta }}.

Unter Verwendung der Legendreschen Verdopplungsformel {displaystyle 2pi Gamma (2eta )={sqrt {pi }}\,Gamma (eta )\,Gamma left(eta +{frac {1}{2}} ight)\,2^{2eta }}

ist das {displaystyle {sqrt {pi }}\,{frac {Gamma left(eta +{frac {1}{2}} ight)}{Gamma (eta )}}\,2^{2eta }\,left(e^{frac {alpha }{2}}+e^{-{frac {alpha }{2}}} ight)^{-2eta }={sqrt {pi }}\,{frac {Gamma left(eta +{frac {1}{2}} ight)}{Gamma (eta )}}\,{ ext{sech}}^{2eta }left({frac {alpha }{2}} ight)}.

 

2.10Bearbeiten

{displaystyle int _{0}^{frac {pi }{2}}cos ^{alpha -1}x\,cos eta x\,dx={frac {pi }{2^{alpha }}}\,{frac {Gamma (alpha )}{Gamma left({frac {alpha +eta +1}{2}} ight)\,Gamma left({frac {alpha -eta +1}{2}} ight)}}qquad { ext{Re}}(alpha )>0\,,\,eta in mathbb {C} }

1. Beweis (Cauchy Cosinus-Integralformel)

In der Formel {displaystyle int _{-infty }^{infty }{frac {dx}{(u+ix)^{a}\,(v-ix)^{b}}}={frac {2pi }{(u+v)^{a+b-1}}}\,{frac {Gamma (a+b-1)}{Gamma (a)\,Gamma (b)}}}

für {displaystyle { ext{Re}}(u),{ ext{Re}}(v)>0;,;{ ext{Re}}(a+b)>1} setze {displaystyle u=v=1\,} und substituiere {displaystyle xmapsto an x}:

{displaystyle I:=int _{-{frac {pi }{2}}}^{frac {pi }{2}}{frac {1}{left(1+i\,{frac {sin x}{cos x}} ight)^{a}\,left(1-i\,{frac {sin x}{cos x}} ight)^{b}}}\,{frac {dx}{cos ^{2}x}}={frac {2pi }{2^{a+b-1}}}\,{frac {Gamma (a+b-1)}{Gamma (a)\,Gamma (b)}}}

Nun ist {displaystyle I=int _{-{frac {pi }{2}}}^{frac {pi }{2}}{frac {(cos x)^{a+b-2}}{(cos x+isin x)^{a}\,(cos x-isin x)^{b}}}\,dx=int _{-{frac {pi }{2}}}^{frac {pi }{2}}(cos x)^{a+b-2}\,e^{-i(a-b)x}\,dx}

aus symmetriegründen gleich {displaystyle int _{-{frac {pi }{2}}}^{frac {pi }{2}}(cos x)^{a+b-2}\,cos(a-b)x\,dx}.

Also ist {displaystyle int _{0}^{frac {pi }{2}}(cos x)^{a+b-2}\,cos(a-b)x\,dx={frac {pi }{2^{a+b-1}}}\,{frac {Gamma (a+b-1)}{Gamma (a)\,Gamma (b)}}}.

Substituiert man {displaystyle alpha =a+b-1,\,eta =a-b},

so ist {displaystyle int _{0}^{frac {pi }{2}}(cos x)^{alpha -1}\,cos eta x\,dx={frac {pi }{2^{alpha }}}\,{frac {Gamma (alpha )}{Gamma left({frac {alpha +eta +1}{2}} ight)\,Gamma left({frac {alpha -eta +1}{2}} ight)}}}.

2. Beweis

Nach der Formel {displaystyle cos 2eta x=sum _{n=0}^{infty }{frac {eta \,(eta -1+n)!}{(eta -n)!}}\,{frac {(2sin x)^{2n}}{(2n)!}}} ist

{displaystyle 2int _{0}^{frac {pi }{2}}cos ^{2alpha -1}x\,cos 2eta x\,dx=sum _{n=0}^{infty }(-1)^{n}\,{frac {eta \,(eta -1+n)!\,2^{2n}}{(eta -n)!\,(2n)!}}\,;2int _{0}^{frac {pi }{2}}sin ^{2n}x\,cos ^{2alpha -1}\,dx}

{displaystyle =sum _{n=0}^{infty }(-1)^{n}\,{frac {eta \,(eta -1+n)!\,2^{2n}}{(eta -n)!\,(2n)!}}\,{frac {left(n-{frac {1}{2}} ight)!\,(alpha -1)!}{left(alpha +n-{frac {1}{2}} ight)!}}}.

Nach der Legendreschen Verdopplungsformel {displaystyle {frac {2^{2n}\,left(n-{frac {1}{2}} ight)!}{(2n)!}}={frac {sqrt {pi }}{n!}}}

ist dies {displaystyle eta \,(alpha -1)!\,{sqrt {pi }}\,sum _{n=0}^{infty }(-1)^{n}\,{frac {(eta -1+n)!}{n!\,(eta -n)!\,left(alpha -{frac {1}{2}}+n ight)!}}}.

Ersetzt man {displaystyle (-1)^{n}\,(eta -1+n)!\,} durch {displaystyle {frac {(eta -1)!\,(-eta )!}{(-eta -n)!}}}, so ist das

{displaystyle eta !\,(-eta )!\,(alpha -1)!\,{sqrt {pi }}\,sum _{n=0}^{infty }{frac {1}{n!\,left(alpha -{frac {1}{2}}+n ight)!\,(eta -n)!\,(-eta -n)!}}}

{displaystyle ={frac {{sqrt {pi }}\,left(alpha -{frac {1}{2}} ight)!\,(alpha -1)!}{left(alpha +eta -{frac {1}{2}} ight)!\,left(alpha -eta -{frac {1}{2}} ight)!}}={frac {pi }{2^{2alpha -1}}}\,{frac {Gamma (2alpha )}{Gamma left(alpha +eta +{frac {1}{2}} ight)\,Gamma left(alpha -eta +{frac {1}{2}} ight)}}}.

Also ist {displaystyle 2int _{0}^{frac {pi }{2}}cos ^{alpha -1}x\,cos eta x\,dx={frac {pi }{2^{alpha }}}\,{frac {Gamma (alpha )}{Gamma left({frac {alpha +eta +1}{2}} ight)\,Gamma left({frac {alpha -eta +1}{2}} ight)}}}.

Beweis für {displaystyle 0<alpha <eta +1}

Es sei {displaystyle H={zin mathbb {C} \,|\,{ ext{Im}}(z)>0}} die obere komplexe Halbebene.

Die Funktion {displaystyle f(z)=(1+z)^{alpha -1}\,z^{eta -1}\,}, mit {displaystyle alpha ,eta >0\,}, ist holomorph auf {displaystyle H\,} und stetig auf {displaystyle {ar {H}}\,}.



Also gilt {displaystyle oint _{K}f\,dz=0}, gleichbedeutend mit {displaystyle underbrace {int _{-1}^{0}f(x)\,dx} _{S_{1}}+underbrace {int _{0}^{1}f(x)\,dx} _{S_{2}}+underbrace {int _{0}^{frac {pi }{2}}f(e^{2ix})\,e^{2ix}\,2i\,dx} _{S_{3}}=0}.

Das erste Integral {displaystyle S_{1}\,} ist nach Substitution {displaystyle xmapsto -x\,} gleich {displaystyle int _{0}^{1}f(-x)\,dx}

{displaystyle =int _{0}^{1}(1-x)^{alpha -1}\,(-x)^{eta -1}\,dx=e^{ipi (eta -1)}int _{0}^{1}(1-x)^{alpha -1}\,x^{eta -1}\,dx=-e^{ipi eta }\,B(alpha ,eta )}.

{displaystyle Rightarrow \,{ ext{Im}}(S_{1})=-sin pi eta \,\,B(alpha ,eta )=-{frac {pi }{Gamma (eta )\,Gamma (1-eta )}}\,{frac {Gamma (alpha )\,Gamma (eta )}{Gamma (alpha +eta )}}=-pi \,{frac {Gamma (alpha )}{Gamma (1-eta )\,Gamma (alpha +eta )}}}.

Das zweite Integral {displaystyle S_{2}\,} ist reell, d.h. {displaystyle { ext{Im}}(S_{2})=0\,}.

Und das dritte Integral {displaystyle S_{3}\,} ist {displaystyle 2i\,int _{0}^{frac {pi }{2}}(1+e^{2ix})^{alpha -1}\,e^{2ix(eta -1)}\,e^{2ix}\,dx}

{displaystyle =2i\,int _{0}^{frac {pi }{2}}(e^{-ix}+e^{ix})^{alpha -1}\,e^{i(alpha +2eta -1)x}\,dx\,Rightarrow \,{ ext{Im}}(S_{3})=2^{alpha }int _{0}^{frac {pi }{2}}cos ^{alpha -1}x\,cos(alpha +2eta -1)x\,dx}.

Aus der Betrachtung der Imaginärteile folgt {displaystyle 2^{alpha }\,int _{0}^{frac {pi }{2}}cos ^{alpha -1}x\,cos(alpha +2eta -1)x\,dx=pi \,{frac {Gamma (alpha )}{Gamma (1-eta )\,Gamma (alpha +eta )}}}.

Ersetzt man {displaystyle alpha +2eta -1\,} durch {displaystyle gamma \,}, also {displaystyle eta ={frac {gamma -alpha +1}{2}}\,}, so ist {displaystyle int _{0}^{frac {pi }{2}}cos ^{alpha -1}x\,cos gamma x\,dx={frac {pi }{2^{alpha }}}\,{frac {Gamma (alpha )}{Gamma left({frac {alpha -gamma +1}{2}} ight)\,Gamma left({frac {alpha +gamma +1}{2}} ight)}}}.

 

3.1Bearbeiten

{displaystyle int _{-infty }^{infty }{frac {cos alpha x}{(eta ^{2}+x^{2})((eta +1)^{2}+x^{2})cdots ((eta +n)^{2}+x^{2})}}\,dx=2pi sum _{k=0}^{n}(-1)^{k}\,{frac {(2eta -1+k)!}{(2eta +n+k)!}}\,{frac {1}{k!\,(n-k)!}}\,e^{-alpha (eta +k)}qquad nin mathbb {N} ;;,;;alpha ,eta >0}

Beweis

Es sei {displaystyle g(z)=prod _{ell =0}^{n}left((eta +ell )^{2}+z^{2} ight)\,,\,f(z)={frac {e^{ialpha z}}{g(z)}}} und {displaystyle gamma _{R}\,} sei der Halbmond in der oberen komplexen Halbebene.

{displaystyle I:=int _{-infty }^{infty }{frac {cos alpha x}{(eta ^{2}+x^{2})((eta +1)^{2}+x^{2})cdots ((eta +n)^{2}+x^{2})}}\,dx}

{displaystyle =lim _{R o infty }oint _{gamma _{R}}f(z)\,dz=2pi i\,sum _{k=0}^{n}{ ext{res}}left(f,i(eta +k) ight)=2pi i\,sum _{k=0}^{n}{frac {e^{-alpha (eta +k)}}{g'left(i(eta +k) ight)}}}.

Nach der Produktregel ist {displaystyle g'(z)=sum _{k=0}^{n}prod _{ell =0}^{k-1}left((eta +ell )^{2}+z^{2} ight)cdot 2zcdot prod _{ell =k+1}^{n}left((eta +ell )^{2}+z^{2} ight)}.

Setzt man {displaystyle z=i(eta +k)\,}, so ist {displaystyle g'left(i(eta +k) ight)=prod _{ell =0}^{k-1}left((eta +ell )^{2}-(eta +k)^{2} ight)cdot 2i(eta +k)cdot prod _{ell =k+1}^{n}left((eta +ell )^{2}-(eta +k)^{2} ight)},

wobei {displaystyle (eta +ell )^{2}-(eta +k)^{2}=(ell -k)(2eta +ell +k)} ist.

Also ist {displaystyle g'left(i(eta +k) ight)=(-1)^{k}\,k!\,{frac {(2eta +2k-1)!}{(2eta -1+k)!}}cdot 2i(eta +k)cdot (n-k)!\,{frac {(2eta +n+k)!}{(2eta +2k)!}}=i\,(-1)^{k}\,{frac {(2eta +n+k)!}{(2eta -1+k)!}}\,k!\,(n-k)!}.

Und somit ist {displaystyle I=2pi sum _{k=0}^{n}(-1)^{k}\,{frac {(2eta -1+k)!}{(2eta +n+k)!}}\,{frac {1}{k!\,(n-k)!}}\,e^{-alpha (eta +k)}}.

 

3.2Bearbeiten

{displaystyle int _{0}^{infty }cos left(alpha \,t^{frac {1}{z}}+eta ight)\,dt={frac {Gamma (z+1)}{alpha ^{z}}}\,cos left({frac {pi z}{2}}+eta ight)qquad 0<z<1\,,\,alpha >0\,,\,eta in mathbb {C} }

ohne Beweis