Zurück zu Unendliche Reihen
S := ∑ n , m = 1 ∞ 1 n m 1 ( n + m ) 2 = ∑ n , m = 1 ∞ 1 n m ∫ 0 1 x n + m d x x ∫ 0 1 y n + m d y y {\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}}} = ∑ n , m = 1 ∞ 1 n m ∫ 0 1 ∫ 0 1 ( x y ) n + m d x d y x y = ∫ 0 1 ∫ 0 1 ∑ n , m = 1 ∞ ( x y ) n n ( x y ) m m d x d y x 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}}} = ∫ 0 1 ∫ 0 1 log 2 ( 1 − x y ) x y d x d y {\displaystyle =\int _{0}^{1}\int _{0}^{1}{\frac {\log ^{2}(1-xy)}{xy}}dx\,dy} Substituiere u = x y {\displaystyle u=xy\,} : S = ∫ 0 1 ∫ 0 y log 2 ( 1 − u ) u d u y d y {\displaystyle S=\int _{0}^{1}\int _{0}^{y}{\frac {\log ^{2}(1-u)}{u}}\,{\frac {du}{y}}\,dy} Und vertausche die Integrationsreihenfolge: S = ∫ 0 1 ∫ u 1 log 2 ( 1 − u ) u d y y d u = ∫ 0 1 − log u u log 2 ( 1 − u ) d u {\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} = ∫ 0 1 − log ( 1 − u ) 1 − u ( log u ) 2 d u = ∫ 0 1 ∑ n = 0 ∞ H n u n ( log u ) 2 d u {\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} = ∑ n = 0 ∞ H n ∫ 0 1 u n ( log u ) 2 d u = ∑ n = 0 ∞ H n 2 ( n + 1 ) 3 = 2 ∑ n = 1 ∞ H n − 1 n 3 {\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}}}} = 2 ( ∑ n = 1 ∞ H n n 3 − ∑ n = 1 ∞ 1 n 4 ) = 2 ( 5 4 ζ ( 4 ) − ζ ( 4 ) ) = 1 2 ζ ( 4 ) {\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 ∑ k = 1 ∞ cos 2 k π x ( k π ) 2 n = ( − 1 ) n − 1 2 2 2 n B 2 n ( x ) ( 2 n ) ! 0 < x < 1 {\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 ∑ k , j , ℓ = 1 ∞ cos 2 k π x ( k π ) 2 n cos 2 j π x ( j π ) 2 n cos 2 ℓ π x ( ℓ π ) 2 n = ( − 1 ) n − 1 8 ( 2 2 n B 2 n ( x ) ( 2 n ) ! ) 3 {\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 x {\displaystyle x\,} von 0 {\displaystyle 0\,} bis 1 {\displaystyle 1\,} . Wegen ∫ 0 1 cos 2 k π x cos 2 j π x cos 2 ℓ π x d x = { 1 4 k + j = ℓ , ℓ + k = j , j + ℓ = k 0 sonst {\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 ∑ k , j = 1 ∞ 1 4 ( k π ) 2 n ( j π ) 2 n ( k π + j π ) 2 n + ∑ ℓ , k = 1 ∞ 1 4 ( ℓ π ) 2 n ( k π ) 2 n ( ℓ π + k π ) 2 n + ∑ j , ℓ = 1 ∞ 1 4 ( j π ) 2 n ( ℓ π ) 2 n ( j π + ℓ π ) 2 n {\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}}}} = 3 4 ∑ k , j = 1 ∞ 1 ( k π ) 2 n ( j π ) 2 n ( k π + j π ) 2 n {\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 ( − 1 ) n − 1 8 ∫ 0 1 ( 2 2 n B 2 n ( x ) ( 2 n ) ! ) 3 d x {\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.
∑ n , m ∈ Z ′ x n 2 + m 2 = ( ∑ n ∈ Z x n 2 ) 2 − 1 = 4 ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 1 − x 2 n + 1 = 4 ∑ n = 0 ∞ ∑ m = 1 ∞ ( − 1 ) n x ( 2 m + 1 ) m {\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 ( − log x ) s − 1 x {\displaystyle {\frac {(-\log x)^{s-1}}{x}}} durch und integriere von 0 bis 1: ∑ n , m ∈ Z ′ ∫ 0 1 x n 2 + m 2 ( − log x ) s − 1 x d x = 4 ∑ n = 0 ∞ ∑ m = 1 ∞ ( − 1 ) n ∫ 0 1 x ( 2 n + 1 ) m ( − log x ) s − 1 x d x {\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} ⇒ ∑ n , m ∈ Z ′ Γ ( s ) ( n 2 + m 2 ) s = 4 ∑ n = 0 ∞ ∑ m = 1 ∞ ( − 1 ) n Γ ( s ) ( ( 2 n + 1 ) m ) s {\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}}}} ⇒ ∑ n , m ∈ Z ′ 1 ( n 2 + m 2 ) s = 4 ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) s ∑ m = 1 ∞ 1 m s = 4 β ( s ) ζ ( 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)}
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