摘要： 0.1Bearbeiten {\displaystyle \int _{0}^{1}\log \Gamma (x)\,dx=\log {\sqrt {2\pi }}} 1. Beweis {\displaystyle 2\int _{0}^{1}\log \Gamma (x)\,dx=\int _{ 阅读全文

摘要： 0.1Bearbeiten {\displaystyle \int _{0}^{1}{\frac {{\text{artanh}}\,x\,\,\log x}{x\,(1-x)\,(1+x)}}\,dx=-{\frac {1}{16}}{\Big (}7\zeta (3)+2\pi ^{2}\log 阅读全文

摘要： 1.1Bearbeiten {\displaystyle \int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2}+x^{2})}{\cosh \pi x}}\,dx=4\log \left({\sqrt {2}}\,\,{\frac {\Gamma \l 阅读全文

摘要： 0.1Bearbeiten {\displaystyle \int _{0}^{1}\log \left(\tan {\frac {\pi x}{2}}\right)\,dx=0} ohne Beweis 0.2Bearbeiten {\displaystyle \int _{0}^{\pi }\l 阅读全文

摘要： 0.1Bearbeiten {\displaystyle \int _{0}^{\pi }\log \left(\cos {\frac {x}{2}}\right)\,dx=-\pi \log 2} Beweis Aus der Fourierreihendarstellung {\displays 阅读全文

摘要： 0.1Bearbeiten {\displaystyle \int _{0}^{\frac {\pi }{2}}\log \left(2\sin {\frac {x}{2}}\right)dx=-G} Beweis Verwende die Fourierreihe {\displaystyle - 阅读全文

摘要： 1.1Bearbeiten {\displaystyle \int _{0}^{\infty }{\text{erf}}^{\;2}\left({\sqrt {x}}\,\right)\,e^{-ax}\,dx={\frac {4}{a\pi }}\cdot {\frac {\operatornam 阅读全文

摘要： 2.1Bearbeiten {\displaystyle {\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }\Gamma (s)\,t^{-s}\,ds=e^{-t}\qquad a>0\,,\,{\text{Re}}(t)>0} Beweis (C 阅读全文

摘要： 1.1Bearbeiten {\displaystyle \int _{0}^{\infty }{\frac {\arctan \left({\frac {x}{z}}\right)}{e^{2\pi x}-1}}\,dx={\frac {1}{2}}\log \left({\frac {z!\,e 阅读全文

摘要： 1.1Bearbeiten {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,\cos(2ax)\,dx={\sqrt {\pi }}\cdot e^{-a^{2}}\qquad a\in \mathbb {C} } 1. Beweis Ver 阅读全文