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

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

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

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

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

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

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

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

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

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

Aus der Kummerschen Reihe

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

folgt

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

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

Also ist

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