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Aus der Partialbruchentwicklung π 2 sec ( π z 2 ) = ∑ k = − ∞ ∞ ( − 1 ) k 2 k + 1 − z {\displaystyle {\frac {\pi }{2}}\,\sec \left({\frac {\pi z}{2}}\right)=\sum _{k=-\infty }^{\infty }{\frac {(-1)^{k}}{2k+1-z}}} ergibt sich π 4 sec ( π z 2 ) = ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 ( 2 k + 1 ) 2 − z 2 {\displaystyle {\frac {\pi }{4}}\,\sec \left({\frac {\pi z}{2}}\right)=\sum _{k=0}^{\infty }(-1)^{k}{\frac {2k+1}{(2k+1)^{2}-z^{2}}}} = ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 ∑ n = 0 ∞ ( z 2 k + 1 ) 2 n = ∑ n = 0 ∞ ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) 2 n + 1 z 2 n {\displaystyle =\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\,\sum _{n=0}^{\infty }\left({\frac {z}{2k+1}}\right)^{2n}=\sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)^{2n+1}}}\,z^{2n}} . Und aus sec ( z ) = ∑ n = 0 ∞ | E 2 n | z 2 n ( 2 n ) ! {\displaystyle \sec(z)=\sum _{n=0}^{\infty }|E_{2n}|\,{\frac {z^{2n}}{(2n)!}}} folgt π 4 sec ( π z 2 ) = ∑ n = 0 ∞ | E 2 n | π 2 n + 1 2 2 n + 2 ( 2 n ) ! z 2 n {\displaystyle {\frac {\pi }{4}}\,\sec \left({\frac {\pi z}{2}}\right)=\sum _{n=0}^{\infty }|E_{2n}|\,{\frac {\pi ^{2n+1}}{2^{2n+2}\,(2n)!}}\,z^{2n}} . Durch Koeffizientenvergleich erhält man ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) 2 n + 1 = | E 2 n | π 2 n + 1 2 2 n + 2 ( 2 n ) ! {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)^{2n+1}}}=|E_{2n}|\,{\frac {\pi ^{2n+1}}{2^{2n+2}\,(2n)!}}} .
Bei der Formel I n := 1 2 | E 2 n | ( π 2 ) 2 n + 1 = ∫ 0 1 log 2 n x 1 + x 2 d x {\displaystyle I_{n}:={\frac {1}{2}}\,|E_{2n}|\,\left({\frac {\pi }{2}}\right)^{2n+1}=\int _{0}^{1}{\frac {\log ^{2n}x}{1+x^{2}}}\,dx} entwickle 1 1 + x 2 {\displaystyle {\frac {1}{1+x^{2}}}} in eine Potenzreihe: I n = ∑ k = 0 ∞ ( − 1 ) k ∫ 0 1 log 2 n x x 2 k d x {\displaystyle I_{n}=\sum _{k=0}^{\infty }(-1)^{k}\int _{0}^{1}\log ^{2n}x\,\,x^{2k}\,dx} ∫ 0 1 log 2 n x x 2 k d x {\displaystyle \int _{0}^{1}\log ^{2n}x\,\,x^{2k}\,dx} ist nach Substitution x = e − t {\displaystyle x=e^{-t}\,} gleich ∫ 0 ∞ t 2 n e − ( 2 k + 1 ) t d t = ( 2 n ) ! ( 2 k + 1 ) 2 n + 1 {\displaystyle \int _{0}^{\infty }t^{2n}\,e^{-(2k+1)t}\,dt={\frac {(2n)!}{(2k+1)^{2n+1}}}} . Also ist ( 2 n ) ! ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) 2 n + 1 = 1 2 | E 2 n | ( π 2 ) 2 n + 1 {\displaystyle (2n)!\,\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)^{2n+1}}}={\frac {1}{2}}\,|E_{2n}|\,\left({\frac {\pi }{2}}\right)^{2n+1}} .
Differenziert man die Formel β ( α ) Γ ( α ) = 1 2 ∫ 0 ∞ x α − 1 cosh x d x {\displaystyle \beta (\alpha )\,\Gamma (\alpha )={\frac {1}{2}}\int _{0}^{\infty }{\frac {x^{\alpha -1}}{\cosh x}}\,dx} , so ist β ′ ( α ) Γ ( α ) + Γ ′ ( α ) β ( α ) = 1 2 ∫ 0 ∞ x α − 1 log x cosh x d x {\displaystyle \beta '(\alpha )\,\Gamma (\alpha )+\Gamma '(\alpha )\,\beta (\alpha )={\frac {1}{2}}\int _{0}^{\infty }{\frac {x^{\alpha -1}\,\log x}{\cosh x}}\,dx} . Setzt man anschließend α = 1 {\displaystyle \alpha =1\,} , so ist β ′ ( 1 ) − γ π 4 = 1 2 ∫ 0 ∞ log x cosh x d x = π 2 log ( 2 π Γ ( 3 4 ) Γ ( 1 4 ) ) {\displaystyle \beta '(1)-\gamma \,{\frac {\pi }{4}}={\frac {1}{2}}\int _{0}^{\infty }{\frac {\log x}{\cosh x}}\,dx={\frac {\pi }{2}}\log \left({\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)} .
In der Legendreschen Verdopplungsformel Γ ( x ) Γ ( x + 1 2 ) = π 2 2 x − 1 Γ ( 2 x ) {\displaystyle \Gamma (x)\,\Gamma \left(x+{\frac {1}{2}}\right)={\frac {\sqrt {\pi }}{2^{2x-1}}}\,\Gamma (2x)} setze x = 1 4 {\displaystyle x={\frac {1}{4}}} , dann ist Γ ( 1 4 ) Γ ( 3 4 ) = 2 π π ⇒ 2 π Γ ( 3 4 ) Γ ( 1 4 ) = 2 π 3 2 Γ 2 ( 1 4 ) {\displaystyle \Gamma \left({\frac {1}{4}}\right)\,\Gamma \left({\frac {3}{4}}\right)={\sqrt {2\pi }}\,{\sqrt {\pi }}\Rightarrow {\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}={\frac {{\sqrt {2\pi }}^{\,3}}{{\sqrt {2}}\,\,\Gamma ^{2}\left({\frac {1}{4}}\right)}}} ⇒ log ( 2 π Γ ( 3 4 ) Γ ( 1 4 ) ) = 3 2 log 2 π − log 2 − 2 log Γ ( 1 4 ) {\displaystyle \Rightarrow \log \left({\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)={\frac {3}{2}}\log 2\pi -\log {\sqrt {2}}-2\log \Gamma \left({\frac {1}{4}}\right)} ⇒ − 1 2 log ( 2 π Γ ( 3 4 ) Γ ( 1 4 ) ) = log Γ ( 1 4 ) − 3 4 log 2 π + 1 2 log 2 {\displaystyle \Rightarrow -{\frac {1}{2}}\log \left({\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)=\log \Gamma \left({\frac {1}{4}}\right)-{\frac {3}{4}}\log 2\pi +{\frac {1}{2}}\log {\sqrt {2}}} Setzt man in der Kummerschen Reihe ∑ n = 1 ∞ log n n π sin 2 n π x = log Γ ( x ) − 1 2 log 2 π + 1 2 log ( 2 sin π x ) + ( γ + log 2 π ) ( x − 1 2 ) {\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)+(\gamma +\log 2\pi )\left(x-{\frac {1}{2}}\right)} für 0 < x < 1 {\displaystyle 0<x<1\,} x = 1 4 {\displaystyle x={\frac {1}{4}}} , so ist ∑ n = 1 ∞ ( − 1 ) n log ( 2 n + 1 ) ( 2 n + 1 ) π = log Γ ( 1 4 ) − 1 2 log 2 π + 1 2 log 2 − 1 4 ( γ + log 2 π ) {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}\,\log(2n+1)}{(2n+1)\pi }}=\log \Gamma \left({\frac {1}{4}}\right)-{\frac {1}{2}}\log 2\pi +{\frac {1}{2}}\log {\sqrt {2}}-{\frac {1}{4}}\left(\gamma +\log 2\pi \right)} = − γ 4 + log Γ ( 1 4 ) − 3 4 log 2 π + 1 2 log 2 {\displaystyle =-{\frac {\gamma }{4}}+\log \Gamma \left({\frac {1}{4}}\right)-{\frac {3}{4}}\log 2\pi +{\frac {1}{2}}\log {\sqrt {2}}} . Also ist ∑ n = 1 ∞ ( − 1 ) n log ( 2 n + 1 ) 2 n + 1 = − π γ 4 − π 2 log ( 2 π Γ ( 3 4 ) Γ ( 1 4 ) ) {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}\,\log(2n+1)}{2n+1}}=-{\frac {\pi \gamma }{4}}-{\frac {\pi }{2}}\log \left({\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)} .
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