• 微信公众号：美女很有趣。 工作之余，放松一下，关注即送10G+美女照片！

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

1周前 (05-04) 5次浏览

##### 1.1Bearbeiten
{displaystyle J_{nu }(z)={frac {2}{Gamma left({frac {1}{2}}right)Gamma left(nu +{frac {1}{2}}right)}}left({frac {z}{2}}right)^{nu }int _{0}^{frac {pi }{2}}cos(zcos x),sin ^{2nu }x,dxqquad {text{Re}}(nu )>-{frac {1}{2}}}
Beweis (Poissonsche Darstellungen der Besselfunktion)

In der Formel

{displaystyle J_{nu }(z)={frac {2}{Gamma left({frac {1}{2}}right)Gamma left(nu +{frac {1}{2}}right)}}left({frac {z}{2}}right)^{nu }int _{0}^{1}cos(zx),(1-x^{2})^{nu -{frac {1}{2}}},dxqquad {text{Re}}(nu )>-{frac {1}{2}}}

substituiere {displaystyle xmapsto cos x}.

##### 1.2Bearbeiten
{displaystyle J_{nu }(z)={frac {2}{Gamma left({frac {1}{2}}right)Gamma left(nu +{frac {1}{2}}right)}}left({frac {z}{2}}right)^{nu }int _{0}^{frac {pi }{2}}cos(zsin x),cos ^{2nu }x,dxqquad {text{Re}}(nu )>-{frac {1}{2}}}
Beweis

In der Formel

{displaystyle J_{nu }(z)={frac {2}{Gamma left({frac {1}{2}}right)Gamma left(nu +{frac {1}{2}}right)}}left({frac {z}{2}}right)^{nu }int _{0}^{1}cos(zx),(1-x^{2})^{nu -{frac {1}{2}}},dxqquad {text{Re}}(nu )>-{frac {1}{2}}}

substituiere {displaystyle xmapsto sin x}.

##### 1.3Bearbeiten
{displaystyle int _{0}^{frac {pi }{2}}{frac {sin alpha x}{sin x}},cos ^{alpha -1}x,dx={frac {pi }{2}}qquad {text{Re}}(alpha )>0}
Beweis (Liouvillesches Integral)

Aus der Formel {displaystyle sin 2alpha x=alpha sum _{n=0}^{infty }(-1)^{n},{frac {left(alpha -{frac {1}{2}}+nright)!}{left(alpha -{frac {1}{2}}-nright)!}},{frac {(2sin x)^{2n+1}}{(2n+1)!}}} für {displaystyle {text{Re}}(alpha )>0,} und {displaystyle 0leq xleq {frac {pi }{2}}} folgt

{displaystyle I:=int _{0}^{frac {pi }{2}}{frac {sin 2alpha x}{sin x}},cos ^{2alpha -1}x,dx=alpha sum _{n=0}^{infty }(-1)^{n},{frac {left(alpha -{frac {1}{2}}+nright)!}{left(alpha -{frac {1}{2}}-nright)!}},{frac {2^{2n}}{(2n)!}},{frac {2}{2n+1}},int _{0}^{frac {pi }{2}}sin ^{2n}x,cos ^{2alpha -1}x,dx},

dabei ist {displaystyle 2int _{0}^{frac {pi }{2}}sin ^{2n}x,cos ^{2alpha -1}x,dx=Bleft(n+{frac {1}{2}},alpha right)={frac {left(n-{frac {1}{2}}right)!,(alpha -1)!}{left(n-{frac {1}{2}}+alpha right)!}}}.

Also ist {displaystyle I=alpha !,sum _{n=0}^{infty }(-1)^{n},{frac {2^{2n},left(n-{frac {1}{2}}right)!}{left(alpha -{frac {1}{2}}-nright)!,(2n)!}},{frac {1}{2n+1}}}.

Nach der Legendreschen Verdopplungsformel lässt sich {displaystyle {frac {2^{2n},left(n-{frac {1}{2}}right)!}{(2n)!}}} durch {displaystyle {frac {sqrt {pi }}{n!}}} ersetzen.

{displaystyle I={sqrt {pi }},alpha !,sum _{n=0}^{infty }{frac {1}{left(alpha -{frac {1}{2}}-nright)!,n!}},{frac {1}{2n+1}}={frac {sqrt {pi }}{2}},{frac {alpha !}{left(alpha -{frac {1}{2}}right)!}},sum _{n=0}^{infty }(-1)^{n},{alpha -{frac {1}{2}} choose n},{frac {1}{n+{frac {1}{2}}}}}.

Nach der Formel {displaystyle sum _{n=0}^{infty }(-1)^{n},{x-1 choose n},{frac {1}{n+y}}=B(x,y)} ist dies {displaystyle {frac {sqrt {pi }}{2}},{frac {alpha !}{left(alpha -{frac {1}{2}}right)!}},Bleft(alpha +{frac {1}{2}},{frac {1}{2}}right)={frac {sqrt {pi }}{2}},Gamma left({frac {1}{2}}right)={frac {pi }{2}}}.

##### 1.4Bearbeiten
{displaystyle int _{0}^{frac {pi }{2}}{frac {sin 2alpha x}{sin x}},cos x,dx={frac {pi }{2}}+{frac {sin alpha pi }{2}}left(psi left({frac {alpha }{2}}right)-psi left({frac {alpha }{2}}+{frac {1}{2}}right)+{frac {1}{alpha }}right)}
ohne Beweis

##### 1.5Bearbeiten
{displaystyle int _{0}^{frac {pi }{2}}{frac {k,cos ,x}{sqrt {1-k^{2}sin ^{2}x}}},dx=arcsin k}
ohne Beweis

##### 1.6Bearbeiten
{displaystyle int _{0}^{frac {pi }{2}}{frac {k,sin x,cos ^{2}x}{sqrt {1-k^{2}sin ^{2}x}}},dx={frac {1}{2k}}+{frac {k^{2}-1}{2k^{2}}},{text{artanh}},k}
ohne Beweis

##### 2.1Bearbeiten
{displaystyle int _{0}^{pi }sin nx,cos mx,dx=left{{begin{matrix}0&,&mequiv nmod 2\\{frac {1}{n+m}}+{frac {1}{n-m}}&,&mathrm {sonst} end{matrix}}right.}
ohne Beweis

##### 2.2Bearbeiten
{displaystyle int _{-pi }^{pi }sin ^{n}x,cos ^{m}x,dx={frac {{frac {n!}{2^{n}left({frac {n}{2}}right)!}},{frac {m!}{2^{m}left({frac {m}{2}}right)!}}}{left({frac {n+m}{2}}right)!}}} wenn {displaystyle n,min mathbb {N} } beide gerade sind, andernfalls ist das Integral 0.
ohne Beweis

##### 2.3Bearbeiten
{displaystyle J_{n}(z)={frac {1}{pi }}int _{0}^{pi }cos(zsin x-nx),dxqquad nin mathbb {Z} ;,;zin mathbb {C} }
1. Beweis (Bessel Integral)

Multipliziere die Jacobi-Anger Entwicklung {displaystyle e^{izsin x}=sum _{nin mathbb {Z} }J_{n}(z),e^{inx}}

mit {displaystyle e^{-imx},} durch und integriere anschließend beide Seiten nach {displaystyle x,} von {displaystyle -pi ,} bis {displaystyle pi ,}:

{displaystyle int _{-pi }^{pi }e^{izsin x},e^{-imx},dx=sum _{nin mathbb {Z} }J_{n}(z)int _{-pi }^{pi }e^{i(n-m)x},dx=sum _{nin mathbb {Z} }J_{n}(z),delta _{nm}cdot 2pi =J_{m}(z)cdot 2pi }

Somit ist {displaystyle int _{-pi }^{pi }cos(zsin x-mx),dx+iint _{-pi }^{pi }sin(zsin x-mx),dx=J_{m}(z)cdot 2pi }.

Also ist {displaystyle 2int _{0}^{pi }cos(zsin x-mx),dx=J_{m}(z)cdot 2pi }.

2. Beweis

Multipliziere die Jacobi-Reihe {displaystyle cos(zsin x)=J_{0}(z)+2sum _{n=1}^{infty }J_{2n}(z)cos 2nx}

mit {displaystyle cos mx,} durch und integriere anschließend beide Seiten nach {displaystyle x,} von {displaystyle 0,} bis {displaystyle pi ,}:

{displaystyle int _{0}^{pi }cos(zsin x),cos mx,dx=J_{0}(z)int _{0}^{pi }cos mx,dx+2sum _{n=1}^{infty }J_{2n}(z)int _{0}^{pi }cos 2nx,cos mx,dx}

{displaystyle =J_{0}(z),delta _{m,0}cdot pi +sum _{n=1}^{infty }J_{2n}(z),delta _{m,2n}cdot pi =left{{begin{matrix}J_{m}(z)cdot pi &m;{text{gerade}}\0&m;{text{ungerade}}end{matrix}}right.}.

Multipliziere die Jacobi-Reihe {displaystyle sin(zsin x)=2sum _{n=0}^{infty }J_{2n+1}(z)sin(2n+1)x}

mit {displaystyle sin mx,} durch und integriere anschließend beide Seiten nach {displaystyle x,} von {displaystyle 0,} bis {displaystyle pi ,}:

{displaystyle int _{0}^{pi }sin(zsin x),sin mx,dx=2sum _{n=0}^{infty }J_{2n+1}(z)int _{0}^{pi }sin(2n+1)x,sin mx,dx}

Also ist {displaystyle J_{m}(z)cdot pi =int _{0}^{pi }cos(zsin x),cos mx,dx+int _{0}^{pi }sin(zsin x),sin mx,dx=int _{0}^{pi }cos(zsin x-mx),dx}.

##### 2.4Bearbeiten
{displaystyle 2int _{0}^{frac {pi }{2}}sin ^{2alpha -1}x,cos ^{2beta -1}x,dx=B(alpha ,beta )qquad {text{Re}}(alpha ),{text{Re}}(beta )>0}
Beweis

In der Formel {displaystyle B(alpha ,beta )=int _{0}^{1}x^{alpha -1}(1-x)^{beta -1},dx} substituiere {displaystyle x,} durch {displaystyle sin ^{2}x,}:

{displaystyle B(alpha ,beta )=int _{0}^{1}(sin ^{2}x)^{alpha -1},(cos ^{2}x)^{beta -1}cdot 2,sin x,cos x,dx=2int _{0}^{frac {pi }{2}}sin ^{2alpha -1}x,cos ^{2beta -1}x,dx}

##### 3.1Bearbeiten
{displaystyle int _{0}^{frac {pi }{2}}{frac {dx}{(a^{2},cos ^{2}x+b^{2},sin ^{2}x)^{n+1}}}={frac {pi }{2ab}},sum _{k=0}^{n}{frac {2k choose k}{(2a)^{2k}}},{frac {2(n-k) choose n-k}{(2b)^{2(n-k)}}}}
ohne Beweis