Benutzer:JWmuench/Mathematik

$\int _{0}^{1}{\frac {\log(1+x)}{1+x^{2}}}\,dx={\tfrac {1}{8}}\pi \log 2$
$\int _{0}^{1}{\frac {\log x}{1-x}}\,dx=-{\tfrac {1}{6}}\pi ^{2}$
$\int _{0}^{1}x^{p}\log x\,dx={\frac {-1}{(1+p)^{2}}}$
$\int _{0}^{\tfrac {\pi }{2}}\log \sin x\,dx={\tfrac {1}{2}}\pi \log 2$

$\log \zeta (s)=\sum _{n=1}^{\infty }{\tfrac {1}{n}}P(ns)$

$2\log {\tfrac {2}{5}}=\sum _{n=1}^{\infty }{\frac {1}{2n-1}}P(4n-2)$

$4\log \pi ={\tfrac {3}{2}}\log 2+2\log 3+{\tfrac {1}{2}}\log 5+\sum _{n=1}^{\infty }{\frac {1}{2n-1}}P(4n-2)+\sum _{n=1}^{\infty }{\frac {1}{n}}P(4n)$

Wobei $P(s)$ die Primzetafunktion und $\zeta (s)$ die Riemannsche Zetafunktion bezeichnet.

$\int _{0}^{\tfrac {1}{2}}\Gamma (1+x)\Gamma (1-x)\,dx={\tfrac {2G}{\pi }}$

Noch zu beweisende Zusammenhänge

${\frac {\pi }{4}}=1-2\sum _{k=1}^{\infty }{\frac {1}{(4k-1)(4k+1)}}$

$\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{(2k-1)^{2n+1}}}={\frac {\pi ^{2n+1}}{2^{2n+2}(2n)!}}|E_{2n}|$

$\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {k}{(k+1)^{2}}}={\frac {\pi ^{2}}{12}}-\log 2$

$\sum _{k=1}^{\infty }{\frac {12k^{2}-1}{k(4k^{2}-1)^{2}}}=2\log 2$

$\sum _{k=1}^{\infty }{\frac {1}{k(2k+1)^{2}}}=4-{\frac {\pi ^{2}}{4}}-2\log 2$

$\sum _{k=1}^{\infty }{\frac {1}{(2k-1)(2k+1)}}={\frac {1}{2}}$

$\sum _{k=1}^{\infty }{\frac {1}{(k-1)(k+1)}}={\frac {3}{4}}$

$\sum _{k=1}^{\infty }{\frac {1}{(2k-1)2k(2k+1)}}=\log 2-{\frac {1}{2}}$

$\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{(2k-1)2k(2k+1)}}={\frac {1}{2}}-{\tfrac {1}{2}}\log 2$

$\sum _{k=1}^{\infty }{\frac {1}{k(2k+1)}}=2-2\log 2$

$\sum _{k=1}^{\infty }{\frac {1}{2^{k}\cdot k^{2}}}={\frac {\pi ^{2}}{12}}-{\tfrac {1}{2}}\log ^{2}2$

$\sum _{k=1}^{\infty }{\frac {k}{(4k^{2}-1)^{2}}}={\tfrac {1}{8}}$

$\sum _{k=1}^{\infty }{\frac {1}{k(4k^{2}-1)^{2}}}={\tfrac {3}{2}}-2\log 2$

$\sum _{k=1}^{\infty }{\frac {1}{k(9k^{2}-1)}}={\tfrac {3}{2}}(\log 3-1)$

$\int _{0}^{\infty }{\frac {\,dx}{1+e^{px}}}={\frac {\log 2}{p}}$

$\int _{0}^{\tfrac {\pi }{4}}x\cdot \tan ^{2}x\,dx={\frac {\pi }{4}}-{\frac {\pi ^{3}}{32}}-{\tfrac {1}{2}}\log 2$

$\int _{0}^{\tfrac {\pi }{4}}x\cdot \tan ^{3}x\,dx={\frac {\pi }{4}}-{\tfrac {1}{2}}+{\frac {\pi }{8}}\log 2-{\tfrac {1}{2}}G$

$\int _{0}^{\tfrac {\pi }{4}}{\frac {x^{2}\cdot \tan x}{\cos ^{2}x}}\,dx={\tfrac {1}{2}}\log 2-{\frac {\pi }{4}}+{\frac {\pi ^{2}}{16}}$

$\int _{0}^{\tfrac {\pi }{4}}\log \tan x\,dx=-\int _{\tfrac {\pi }{4}}^{\tfrac {\pi }{2}}\log \tan x\,dx=-G$

$\int _{0}^{1}{\frac {1+x}{1-x}}\log x\,dx=1-{\frac {\pi ^{2}}{3}}$

$\int _{0}^{1}{\frac {\log x}{(1+x)^{2}}}\,dx=-\log 2$

$\int _{0}^{1}{\frac {\log x}{1+x^{2}}}\,dx=-\int _{0}^{\infty }{\frac {\log x}{1+x^{2}}}\,dx=-G$

$\int _{0}^{1}\log(1-x^{2}){\frac {\,dx}{x}}=-{\frac {\pi ^{2}}{12}}$

$\int _{0}^{1}\log(1-x^{2}){\frac {\,dx}{1+x^{2}}}={\frac {\pi }{4}}\log 2-G$

$\int _{0}^{1}\log({\tfrac {1+x^{2}}{x}}){\frac {\,dx}{1+x^{2}}}={\frac {\pi }{2}}\log 2$

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