Formelsammlung Mathematik: Unendliche Reihen: Kummersche Reihe

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Kummersche Reihe

$\sum _{n=1}^{\infty }{\frac {\log n}{n\pi }}\,\sin(2n\pi x)=\log \Gamma (x)-{\frac {1}{2}}\log 2\pi +{\frac {1}{2}}\log(2\,\sin \pi x)+\left(\gamma +\log 2\pi \right)\left(x-{\frac {1}{2}}\right)\qquad 0<x<1$

Beweis

Die Funktion $f(x)=\log \Gamma (x)\,$ soll in eine Fourierreihe ${\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left(a_{n}\cos n\omega x+b_{n}\sin n\omega x\right)$

mit Periodenlänge $T=1\,$, und somit Kreisfrequenz $\omega ={\frac {2\pi }{T}}=2\pi$, entwickelt werden.

Berechne hierzu die Koeffizienten $a_{n},b_{n}\,$ mit Hilfe der Eulerschen Formeln

$a_{n}={\frac {2}{T}}\int _{0}^{T}f(x)\,\cos n\omega x\,dx$ und $b_{n}={\frac {2}{T}}\int _{0}^{T}f(x)\,\sin n\omega x\,dx$.

${\frac {a_{0}}{2}}=\int _{0}^{1}\log \Gamma (x)\,dx={\frac {1}{2}}\log 2\pi$ und für $n\geq 1$ ist $a_{n}=2\int _{0}^{1}\log \Gamma (x)\,\cos 2\pi nx\,dx={\frac {1}{2n}}$

und $b_{n}=2\int _{0}^{1}\log \Gamma (x)\,\sin 2\pi nx\,dx={\frac {\gamma +\log 2\pi +\log n}{n\pi }}$.

Also gilt $\log \Gamma (x)={\frac {1}{2}}\log 2\pi +\sum _{n=1}^{\infty }{\frac {\cos(n\,2\pi x)}{2n}}+\sum _{n=1}^{\infty }{\frac {\gamma +\log 2\pi +\log n}{n\pi }}\,\sin(n\,2\pi x)$.

Dabei ist $\sum _{n=1}^{\infty }{\frac {\cos(n\,2\pi x)}{2n}}=-{\frac {1}{2}}\log(2\,\sin \pi x)$ und $\left(\gamma +\log 2\pi \right)\sum _{n=1}^{\infty }{\frac {\sin(n\,2\pi x)}{n\pi }}=\left(\gamma +\log 2\pi \right)\left({\frac {1}{2}}-x\right)$.

Daher gilt $\log \Gamma (x)={\frac {1}{2}}\log 2\pi -{\frac {1}{2}}\log(2\,\sin \pi x)+\left(\gamma +\log 2\pi \right)\left({\frac {1}{2}}-x\right)+\sum _{n=1}^{\infty }{\frac {\log n}{n\pi }}\,\sin(2n\pi x)$,

gleichbedeutend mit $\sum _{n=1}^{\infty }{\frac {\log n}{n\pi }}\,\sin(2n\pi x)=\log \Gamma (x)-{\frac {1}{2}}\log 2\pi +{\frac {1}{2}}\log(2\,\sin \pi x)+\left(\gamma +\log 2\pi \right)\left(x-{\frac {1}{2}}\right)$.

 

Malmstén Reihe

$\log {\frac {\Gamma \left({\frac {1}{2}}+x\right)}{\Gamma \left({\frac {1}{2}}-x\right)}}=-2x\,(\gamma +\log 2\pi )+{\frac {2}{\pi }}\sum _{n=1}^{\infty }(-1)^{n}\,{\frac {\log n}{n}}\,\sin(2\pi nx)\qquad -{\frac {1}{2}}<x<{\frac {1}{2}}$

Beweis

Aus der Kummerschen Reihe

$\log \Gamma (x)=-\left(x-{\frac {1}{2}}\right)(\gamma +\log 2\pi )-{\frac {1}{2}}\log(2\sin \pi x)+\log {\sqrt {2\pi }}+{\frac {1}{\pi }}\sum _{n=1}^{\infty }{\frac {\log n}{n}}\,\sin(2n\pi x)$    für    $0<x<1\,$

folgt

$\log \Gamma \left({\frac {1}{2}}+x\right)=-x\,(\gamma +\log 2\pi )-{\frac {1}{2}}\log(2\cos \pi x)+\log {\sqrt {2\pi }}+{\frac {1}{\pi }}\sum _{n=1}^{\infty }(-1)^{n}\,{\frac {\log n}{n}}\,\sin(2n\pi x)$

$\log \Gamma \left({\frac {1}{2}}-x\right)=x\,(\gamma +\log 2\pi )-{\frac {1}{2}}\log(2\cos \pi x)+\log {\sqrt {2\pi }}-{\frac {1}{\pi }}\sum _{n=1}^{\infty }(-1)^{n}\,{\frac {\log n}{n}}\,\sin(2n\pi x)$    für    $-{\frac {1}{2}}<x<{\frac {1}{2}}$.

Also ist

$\log {\frac {\Gamma \left({\frac {1}{2}}+x\right)}{\Gamma \left({\frac {1}{2}}-x\right)}}=-2x\,(\gamma +\log 2\pi )+{\frac {2}{\pi }}\sum _{n=1}^{\infty }(-1)^{n}\,{\frac {\log n}{n}}\,\sin(2\pi nx)\qquad -{\frac {1}{2}}<x<{\frac {1}{2}}$