${\displaystyle \sum _{k=0}^{\infty }(\varrho \,e^{i\varphi })^{k}={\frac {1}{1-\varrho \,e^{i\varphi }}}={\frac {1-\varrho \,e^{-i\varphi }}{(1-\varrho \,e^{i\varphi })(1-\varrho \,e^{-i\varphi })}}={\frac {1-\varrho \cos \varphi +i\varrho \sin \varphi }{1-2\varrho \cos \varphi +\varrho ^{2}}}}$

Der Vergleich von Real- und Imaginärteil liefert die Behauptung.

${\displaystyle \sum _{k=1}^{\infty }{\frac {(\varrho \,e^{i\varphi })^{k}}{k}}=-\log(1-\varrho \,e^{i\varphi })=-\log(1-\varrho \cos \varphi -i\varrho \sin \varphi )}$

${\displaystyle =-{\frac {1}{2}}\log \left((1-\varrho \cos \varphi )^{2}+(\varrho \sin \varphi )^{2}\right)-i\arctan \left({\frac {-\varrho \sin \varphi }{1-\varrho \cos \varphi }}\right)}$

${\displaystyle =-{\frac {1}{2}}\log \left(1-2\varrho \cos \varphi +\varrho ^{2}\right)+i\arctan \left({\frac {\varrho \sin \varphi }{1-\varrho \cos \varphi }}\right)}$

Der Vergleich von Real- und Imaginärteil liefert die Behauptung.

${\displaystyle -\sum _{k=1}^{\infty }{\frac {(-e^{i\varphi })^{k}}{k}}=\log \left(1+e^{i\varphi }\right)=\log \left(1+\cos \varphi +i\sin \varphi \right)}$

${\displaystyle ={\frac {1}{2}}\log \left((1+\cos \varphi )^{2}+\sin ^{2}\varphi \right)+i\arctan \left({\frac {\sin \varphi }{1+\cos \varphi }}\right)}$

${\displaystyle ={\frac {1}{2}}\log(2+2\cos \varphi )+i\arctan \left(\tan {\frac {\varphi }{2}}\right)={\frac {1}{2}}\log \left(4\cos ^{2}{\frac {\varphi }{2}}\right)+i\,{\frac {\varphi }{2}}}$

Der Vergleich von Real- und Imaginärteil liefert die Behauptung.

In den Formeln

${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}\cos k\varphi }{k}}=-\log \left(2\,\cos {\frac {\varphi }{2}}\right)\qquad \sum _{k=1}^{\infty }{\frac {(-1)^{k}\sin k\varphi }{k}}=-{\frac {\varphi }{2}}\qquad -\pi <\varphi <\pi }$

ersetze ${\displaystyle \varphi }$ durch ${\displaystyle \varphi -\pi }$.

${\displaystyle \sum _{k=0}^{\infty }{\frac {\cos(2k+1)\varphi }{2k+1}}={\frac {1}{2}}\left[\sum _{k=1}^{\infty }{\frac {\cos k\varphi }{k}}-\sum _{k=1}^{\infty }{\frac {(-1)^{k}\cos k\varphi }{k}}\right]}$

${\displaystyle ={\frac {1}{2}}\left[-\log \left(2\sin {\frac {\varphi }{2}}\right)+\log \left(2\cos {\frac {\varphi }{2}}\right)\right]=-{\frac {1}{2}}\log \left(\tan {\frac {\varphi }{2}}\right)}$


${\displaystyle \sum _{k=0}^{\infty }{\frac {\sin(2k+1)\varphi }{2k+1}}={\frac {1}{2}}\left[\sum _{k=1}^{\infty }{\frac {\sin k\varphi }{k}}-\sum _{k=1}^{\infty }{\frac {(-1)^{k}\sin k\varphi }{k}}\right]={\frac {1}{2}}\left({\frac {\pi -\varphi }{2}}-{\frac {-\varphi }{2}}\right)={\frac {\pi }{4}}}$

In der Formeln

${\displaystyle \sum _{k=0}^{\infty }{\frac {\cos(2k+1)\varphi }{2k+1}}=-{\frac {1}{2}}\log \left(\tan {\frac {\varphi }{2}}\right)\qquad \sum _{k=0}^{\infty }{\frac {\sin(2k+1)\varphi }{2k+1}}={\frac {\pi }{4}}\qquad 0<\varphi <\pi }$

ersetze ${\displaystyle \varphi }$ durch ${\displaystyle \varphi +{\frac {\pi }{2}}}$.

Es sei ${\displaystyle f(z)=e^{ixz}\,{\frac {\pi }{\sin \pi z}}\,{\frac {1}{z+\alpha }}}$ und ${\displaystyle \gamma _{n}\,}$ das Quadrat mit den Ecken ${\displaystyle (\pm 1\pm i)\left(n+{\frac {1}{2}}\right)}$.

${\displaystyle \lim _{n\to \infty }\oint _{\gamma _{n}}fdz=0}$, da wegen ${\displaystyle |f(z)|^{2}=\left|{\frac {e^{ix(u+iv)}}{\sin \left(\pi (u+iv)\right)}}\right|^{2}={\frac {e^{-2xv}}{\sin ^{2}(\pi u)+\sinh ^{2}(\pi v)}}}$

${\displaystyle f\,}$ asymptotisch abklingt wie ${\displaystyle e^{-xv}\,e^{-\pi |v|}\,{\frac {1}{n}}}$.

${\displaystyle {\text{res}}(f,k)=e^{ikx}\,{\frac {1}{\cos k\pi }}\,{\frac {1}{k+\alpha }}}$ und ${\displaystyle {\text{res}}(f,-\alpha )=e^{-i\alpha x}{\frac {\pi }{-\sin(\alpha \pi )}}}$.

Da die Summe aller Residuen null ergeben muss, ist ${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {(-1)^{k}\,e^{ikx}}{k+\alpha }}=\pi {\frac {e^{-i\alpha x}}{\sin \alpha \pi }}}$.

Für ${\displaystyle 0<x<2\pi \,}$ und ${\displaystyle \alpha \in \mathbb {C} \setminus i\mathbb {R} }$ ist ${\displaystyle S:=\sum _{k\in \mathbb {Z} }{\frac {e^{ikx}}{k-i\alpha }}}$

nach der Poissonschen Summationsformel ${\displaystyle \sum _{k\in \mathbb {Z} }\int _{-\infty }^{\infty }{\frac {e^{itx}}{t-i\alpha }}\,e^{2\pi ikt}\,dt}$.

Und das ist ${\displaystyle \sum _{k=-\infty }^{-1}\oint \limits _{{\text{unterer}} \atop {\text{Halbmond}}}{\frac {e^{ixz}}{z-i\alpha }}\,e^{2\pi ikz}\,dz+\sum _{k=0}^{\infty }\oint \limits _{{\text{oberer}} \atop {\text{Halbmond}}}{\frac {e^{ixz}}{z-i\alpha }}\,e^{2\pi ikz}\,dz}$.

Wegen ${\displaystyle \oint \limits _{{\text{unterer}} \atop {\text{Halbmond}}}...=\left\{{\begin{matrix}-2\pi i\,e^{-\alpha x}\,e^{-2\pi k\alpha }&,&{\text{Re}}(\alpha )<0\\0&,&{\text{Re}}(\alpha )>0\end{matrix}}\right.}$

und ${\displaystyle \oint \limits _{{\text{oberer}} \atop {\text{Halbmond}}}...=\left\{{\begin{matrix}0&,&{\text{Re}}(\alpha )<0\\2\pi i\,e^{-\alpha x}\,e^{-2\pi k\alpha }&,&{\text{Re}}(\alpha )>0\end{matrix}}\right.}$

ist für ${\displaystyle {\text{Re}}(\alpha )<0\qquad S=-2\pi i\,e^{-\alpha x}\,\sum _{k=1}^{\infty }e^{2\pi \alpha k}=2\pi i\,{\frac {e^{-\alpha x}}{1-e^{-2\pi \alpha }}}}$

und für ${\displaystyle {\text{Re}}(\alpha )>0\qquad S=2\pi i\,e^{-\alpha x}\,\sum _{k=0}^{\infty }e^{-2\pi \alpha k}=2\pi i\,{\frac {e^{-\alpha x}}{1-e^{-2\pi \alpha }}}}$ .

Daher ist für ${\displaystyle 0<x<2\pi \,}$ und ${\displaystyle \alpha \in \mathbb {C} \setminus \mathbb {R} }$

${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {e^{ikx}}{k+\alpha }}=2\pi i\,{\frac {e^{-i\alpha x}}{1-e^{-2\pi i\alpha }}}=\pi \,{\frac {e^{-i(x-\pi )\alpha }}{\sin \alpha \pi }}}$.

Verschiebt man ${\displaystyle x\,}$ nach ${\displaystyle x+\pi \,}$, so ist ${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {(-1)^{k}e^{ikx}}{k+\alpha }}=\pi \,{\frac {e^{-i\alpha x}}{\sin \alpha \pi }}}$ für ${\displaystyle -\pi <x<\pi \,}$.

Schreibe ${\displaystyle {\frac {1}{k^{2}+\alpha ^{2}}}}$ als ${\displaystyle {\frac {i}{2\alpha }}\left({\frac {1}{k+i\alpha }}-{\frac {1}{k-i\alpha }}\right)}$

und verwende die Formel ${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {(-1)^{k}\,\cos kx}{k+\alpha }}=\pi \,{\frac {\cos \alpha x}{\sin \alpha \pi }}}$.

Dann ist ${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {(-1)^{k}\,\cos(kx)}{k^{2}+\alpha ^{2}}}={\frac {i\pi }{2\alpha }}\left({\frac {\cosh \alpha x}{i\sinh \alpha \pi }}-{\frac {\cosh \alpha x}{-i\sinh \alpha \pi }}\right)={\frac {\pi }{\alpha }}\,{\frac {\cosh(\alpha x)}{\sinh(\alpha \pi )}}}$.

Ist ${\displaystyle \alpha >0\,}$ und ${\displaystyle 0<x<2\pi \,}$, so ist nach der Poissonschen Summationsformel

${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {e^{ikx}}{k^{2}+\alpha ^{2}}}=\sum _{k\in \mathbb {Z} }\int _{-\infty }^{\infty }{\frac {e^{itx}}{t^{2}+\alpha ^{2}}}\,e^{2\pi ikt}\,dt}$

und das ist ${\displaystyle {\frac {\pi }{\alpha }}\,\sum _{k\in \mathbb {Z} }e^{-\alpha |2\pi k+x|}={\frac {\pi }{\alpha }}\left(\sum _{k=-\infty }^{-1}e^{\alpha (2\pi k+x)}+\sum _{k=0}^{\infty }e^{-\alpha (2\pi k+x)}\right)}$

${\displaystyle ={\frac {\pi }{\alpha }}\left(e^{\alpha x}\sum _{k=1}^{\infty }e^{-2\pi \alpha k}+e^{-\alpha x}\sum _{k=0}^{\infty }e^{-2\pi \alpha k}\right)={\frac {\pi }{\alpha }}\left({\frac {e^{\alpha x}}{e^{2\pi \alpha }-1}}+{\frac {e^{-\alpha x}}{1-e^{-2\pi \alpha }}}\right)}$

${\displaystyle ={\frac {\pi }{\alpha }}\,{\frac {e^{\alpha (x-\pi )}+e^{-\alpha (x-\pi )}}{e^{\pi \alpha }-e^{-\pi \alpha }}}}$. Also ist ${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {\cos(kx)}{k^{2}+\alpha ^{2}}}={\frac {\pi }{\alpha }}\,{\frac {\cosh {\big (}\alpha (x-\pi ){\big )}}{\sinh(\alpha \pi )}}}$

Nach Substitution ${\displaystyle x\mapsto x+\pi }$ erhält man die gesuchte Formel.

Die Funktion ${\displaystyle f(x)=|\sin x|\,}$ hat Periodenlänge ${\displaystyle T=\pi \,}$, und somit Kreisfrequenz ${\displaystyle \omega ={\frac {2\pi }{T}}=2}$.

Die Koeffizienten ${\displaystyle a_{n},b_{n}\,}$ der Fourierreihenentwicklung ${\displaystyle {\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left(a_{n}\cos 2nx+b_{n}\sin 2nx\right)}$ erhält man durch die Eulerschen Formeln:

${\displaystyle a_{n}={\frac {2}{\pi }}\int _{0}^{\pi }|\sin x|\,\cos 2nx\,dx={\frac {2}{\pi }}\left({\frac {1}{2n+1}}-{\frac {1}{2n-1}}\right)\quad ,\quad b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }|\sin x|\,\sin 2nx\,dx=0}$

Also ist ${\displaystyle |\sin x|={\frac {2}{\pi }}+{\frac {2}{\pi }}\sum _{n=1}^{\infty }\left({\frac {1}{2n+1}}-{\frac {1}{2n-1}}\right)\,\cos 2nx}$.

In der Formel ${\displaystyle {\frac {2}{\pi }}+{\frac {2}{\pi }}\sum _{n=1}^{\infty }\left({\frac {1}{2n+1}}-{\frac {1}{2n-1}}\right)\,\cos 2nx=|\sin x|}$ ersetze ${\displaystyle x\,}$ durch ${\displaystyle x+{\frac {\pi }{2}}}$.

Die Funktion ${\displaystyle f\,}$ lässt sich als komplexe Fourierreihe ${\displaystyle f(x)=\sum _{k=-\infty }^{\infty }c_{k}\,e^{ikx}}$ schreiben, wenn man ${\displaystyle 2c_{k}=\left\{{\begin{array}{ll}a_{k}-ib_{k}&k>0\\a_{0}&k=0\\a_{k}+ib_{k}&k<0\end{array}}\right.}$     setzt.

Nun ist ${\displaystyle |f(x)|^{2}=f(x)\,{\overline {f(x)}}=\sum _{k=-\infty }^{\infty }c_{k}\,e^{ikx}\,\sum _{n=-\infty }^{\infty }{\overline {c_{n}}}\,e^{-inx}=\sum _{k,n=-\infty }^{\infty }c_{k}\,{\overline {c_{n}}}\,e^{i(k-n)x}}$, und somit ist

${\displaystyle {\frac {1}{\pi }}\int _{-\pi }^{\pi }|f(x)|^{2}\,dx={\frac {1}{\pi }}\sum _{k,n=-\infty }^{\infty }c_{k}\,{\overline {c_{n}}}\,\underbrace {\int _{-\pi }^{\pi }e^{i(k-n)x}\,dx} _{=2\pi \,\delta _{n,k}}=2\sum _{k=-\infty }^{\infty }|c_{k}|^{2}={\frac {1}{2}}\sum _{k=-\infty }^{\infty }|2c_{k}|^{2}}$

${\displaystyle ={\frac {1}{2}}\left(\sum _{k=-\infty }^{-1}|a_{k}+ib_{k}|^{2}+|a_{0}|^{2}+\sum _{k=1}^{\infty }|a_{k}-ib_{k}|^{2}\right)={\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{\infty }\left(a_{k}^{2}+b_{k}^{2}\right)}$.