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

摘要： 1.1Bearbeiten {\displaystyle \int _{0}^{\infty }\left({\frac {\sin x}{x}}\right)^{2}\,e^{-2ax}\,dx=a\,\log \left({\frac {a}{\sqrt {1+a^{2}}}}\right)+\ 阅读全文

摘要： 0.1Bearbeiten {\displaystyle \int _{0}^{\infty }W\left({\frac {1}{x^{2}}}\right)dx={\sqrt {2\pi }}} Beweis In der Formel {\displaystyle \int _{0}^{\in 阅读全文

摘要： 2.1Bearbeiten {\displaystyle \int _{0}^{\infty }{\text{Ci}}(ax)\,{\text{Ci}}(bx)\,dx={\frac {1}{\max\{a,b\}}}\cdot {\frac {\pi }{2}}\qquad a,b>0} Bewe 阅读全文

摘要： 1.1Bearbeiten {\displaystyle \int _{0}^{\infty }{\frac {x}{\sqrt {\alpha ^{2}+x^{2}}}}\,J_{0}(x)\,dx=e^{-\alpha }\qquad {\text{Re}}(\alpha )\geq 0} Be 阅读全文

摘要： 3.1Bearbeiten {\displaystyle \int _{-\infty }^{\infty }(\alpha -ix)^{n}\,\Gamma (\beta +ix)\,dx={\frac {2\pi }{e}}\sum _{k=0}^{n}{n \choose k}\,(\alph 阅读全文

摘要： 2.1Bearbeiten {\displaystyle \int _{0}^{\infty }{\text{arccot}}(ax)\cdot {\text{arccot}}(bx)\ dx={\frac {\pi }{2}}\left[{\frac {1}{a}}\log \left({\fra 阅读全文

摘要： 0.1Bearbeiten {\displaystyle \int _{0}^{1}{\frac {\arctan x}{x}}\,dx=G} Beweis Benutze die Reihenentwicklung {\displaystyle \arctan x=\sum _{k=0}^{\in 阅读全文

摘要： 0.1Bearbeiten {\displaystyle \int _{0}^{1}{\frac {\arcsin x}{x}}\,dx={\frac {\pi }{2}}\,\log 2} Beweis {\displaystyle \int _{0}^{1}{\frac {\arcsin x}{ 阅读全文

摘要： 0.1Bearbeiten {\displaystyle \int _{-\infty }^{\infty }{\frac {1}{1+x^{2}}}\,{\frac {1}{\cosh {\frac {\pi x}{2}}}}\,dx=2\log 2} ohne Beweis 1.1Bearbei 阅读全文