${\displaystyle S:=\sum _{n,m=1}^{\infty }{\frac {1}{nm}}\,{\frac {1}{(n+m)^{2}}}=\sum _{n,m=1}^{\infty }{\frac {1}{nm}}\,\int _{0}^{1}x^{n+m}\,{\frac {dx}{x}}\,\,\int _{0}^{1}y^{n+m}\,{\frac {dy}{y}}}$

${\displaystyle =\sum _{n,m=1}^{\infty }{\frac {1}{nm}}\int _{0}^{1}\int _{0}^{1}(xy)^{n+m}\,{\frac {dx\,dy}{xy}}=\int _{0}^{1}\int _{0}^{1}\sum _{n,m=1}^{\infty }{\frac {(xy)^{n}}{n}}\,{\frac {(xy)^{m}}{m}}\,{\frac {dx\,dy}{xy}}}$

${\displaystyle =\int _{0}^{1}\int _{0}^{1}{\frac {\log ^{2}(1-xy)}{xy}}dx\,dy}$

Substituiere ${\displaystyle u=xy\,}$:

${\displaystyle S=\int _{0}^{1}\int _{0}^{y}{\frac {\log ^{2}(1-u)}{u}}\,{\frac {du}{y}}\,dy}$

Und vertausche die Integrationsreihenfolge:

${\displaystyle S=\int _{0}^{1}\int _{u}^{1}{\frac {\log ^{2}(1-u)}{u}}\,{\frac {dy}{y}}\,du=\int _{0}^{1}{\frac {-\log u}{u}}\,\log ^{2}(1-u)\,du}$

${\displaystyle =\int _{0}^{1}-{\frac {\log(1-u)}{1-u}}\,(\log u)^{2}\,du=\int _{0}^{1}\sum _{n=0}^{\infty }H_{n}u^{n}\,(\log u)^{2}\,du}$

${\displaystyle =\sum _{n=0}^{\infty }H_{n}\int _{0}^{1}u^{n}\,(\log u)^{2}\,du=\sum _{n=0}^{\infty }H_{n}\,{\frac {2}{(n+1)^{3}}}=2\sum _{n=1}^{\infty }{\frac {H_{n-1}}{n^{3}}}}$

${\displaystyle =2\left(\sum _{n=1}^{\infty }{\frac {H_{n}}{n^{3}}}-\sum _{n=1}^{\infty }{\frac {1}{n^{4}}}\right)=2\left({\frac {5}{4}}\zeta (4)-\zeta (4)\right)={\frac {1}{2}}\zeta (4)}$

Aus der Formel ${\displaystyle \sum _{k=1}^{\infty }{\frac {\cos 2k\pi x}{(k\pi )^{2n}}}={\frac {(-1)^{n-1}}{2}}\,{\frac {2^{2n}\,B_{2n}(x)}{(2n)!}}\qquad 0<x<1}$

folgt ${\displaystyle \sum _{k,j,\ell =1}^{\infty }{\frac {\cos 2k\pi x}{(k\pi )^{2n}}}\,{\frac {\cos 2j\pi x}{(j\pi )^{2n}}}\,{\frac {\cos 2\ell \pi x}{(\ell \pi )^{2n}}}={\frac {(-1)^{n-1}}{8}}\left({\frac {2^{2n}\,B_{2n}(x)}{(2n)!}}\right)^{3}}$.

Integriere nun beide Seiten nach ${\displaystyle x\,}$ von ${\displaystyle 0\,}$ bis ${\displaystyle 1\,}$.

Wegen ${\displaystyle \int _{0}^{1}\cos 2k\pi x\,\cos 2j\pi x\,\cos 2\ell \pi x\,dx=\left\{{\begin{matrix}{\frac {1}{4}}&k+j=\ell \,,\,\ell +k=j\,,\,j+\ell =k\\\\0&{\text{sonst}}\end{matrix}}\right.}$

steht dann auf der linken Seite

${\displaystyle \sum _{k,j=1}^{\infty }{\frac {\frac {1}{4}}{(k\pi )^{2n}\,(j\pi )^{2n}\,(k\pi +j\pi )^{2n}}}+\sum _{\ell ,k=1}^{\infty }{\frac {\frac {1}{4}}{(\ell \pi )^{2n}\,(k\pi )^{2n}\,(\ell \pi +k\pi )^{2n}}}+\sum _{j,\ell =1}^{\infty }{\frac {\frac {1}{4}}{(j\pi )^{2n}\,(\ell \pi )^{2n}\,(j\pi +\ell \pi )^{2n}}}}$

${\displaystyle ={\frac {3}{4}}\,\sum _{k,j=1}^{\infty }{\frac {1}{(k\pi )^{2n}\,(j\pi )^{2n}\,(k\pi +j\pi )^{2n}}}}$, und auf der rechten Seite steht ${\displaystyle {\frac {(-1)^{n-1}}{8}}\int _{0}^{1}\left({\frac {2^{2n}\,B_{2n}(x)}{(2n)!}}\right)^{3}dx}$.

Daraus folgt unmittelbar die behauptete Gleichung.

${\displaystyle {\sum _{n,m\in \mathbb {Z} }}'x^{n^{2}+m^{2}}=\left(\sum _{n\in \mathbb {Z} }x^{n^{2}}\right)^{2}-1=4\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{1-x^{2n+1}}}=4\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }(-1)^{n}\,x^{(2m+1)m}}$

Multipliziere beide Seiten mit ${\displaystyle {\frac {(-\log x)^{s-1}}{x}}}$ durch und integriere von 0 bis 1:

${\displaystyle {\sum _{n,m\in \mathbb {Z} }}'\int _{0}^{1}x^{n^{2}+m^{2}}\,{\frac {(-\log x)^{s-1}}{x}}\,dx=4\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }(-1)^{n}\,\int _{0}^{1}x^{(2n+1)m}\,{\frac {(-\log x)^{s-1}}{x}}\,dx}$

${\displaystyle \Rightarrow \,{\sum _{n,m\in \mathbb {Z} }}'{\frac {\Gamma (s)}{(n^{2}+m^{2})^{s}}}=4\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }(-1)^{n}\,{\frac {\Gamma (s)}{((2n+1)m)^{s}}}}$

${\displaystyle \Rightarrow \,{\sum _{n,m\in \mathbb {Z} }}'{\frac {1}{(n^{2}+m^{2})^{s}}}=4\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{s}}}\,\sum _{m=1}^{\infty }{\frac {1}{m^{s}}}=4\,\beta (s)\,\zeta (s)}$