# Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,arctan)

Zurück zu Bestimmte Integrale

##### 0.1
${\displaystyle \int _{0}^{1}{\frac {\arctan x}{x}}\,dx=G}$
Beweis

Benutze die Reihenentwicklung ${\displaystyle \arctan x=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {x^{2k+1}}{2k+1}}}$.

${\displaystyle \int _{0}^{1}{\frac {\arctan x}{x}}\,dx=\sum _{k=0}^{\infty }(-1)^{k}\int _{0}^{1}{\frac {x^{2k}}{2k+1}}\,dx=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {1}{(2k+1)^{2}}}=G}$

##### 0.2
${\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1+x^{2}}}\,dx={\frac {\pi ^{2}}{8}}}$
Beweis

${\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1+x^{2}}}\,dx=\left[{\frac {1}{2}}\arctan ^{2}x\right]_{0}^{\infty }={\frac {1}{2}}\,\left({\frac {\pi }{2}}\right)^{2}={\frac {\pi ^{2}}{8}}}$

##### 0.3
${\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1-x^{2}}}\,dx=-G}$
ohne Beweis

##### 0.4
${\displaystyle \int _{0}^{\infty }{\frac {x\,\arctan x}{1+x^{4}}}\,dx={\frac {\pi ^{2}}{16}}}$
ohne Beweis

##### 0.5
${\displaystyle \int _{0}^{\infty }{\frac {x\,\arctan x}{1-x^{4}}}\,dx=-{\frac {\pi }{8}}\,\log 2}$
ohne Beweis

##### 0.6
${\displaystyle \int _{0}^{1}{\frac {\arctan \left(x^{3+{\sqrt {8}}\,}\right)}{1+x^{2}}}\,dx={\frac {1}{8}}\,\log(2)\cdot \log \left(1+{\sqrt {2}}\,\right)}$
ohne Beweis

##### 1.1
${\displaystyle \int _{-\infty }^{\infty }{\frac {\arctan ax}{x\,(1+x^{2})}}\,dx=\pi \log(1+a)\qquad a\geq 0}$
Beweis

Integriere die Formel ${\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{(1+t^{2}x^{2})(1+x^{2})}}={\frac {\pi }{1+t}}}$ nach ${\displaystyle t\,}$ von ${\displaystyle 0\,}$ bis ${\displaystyle a\,}$.

##### 1.2
${\displaystyle \int _{0}^{\infty }{\frac {\arctan ax}{x\,(1-x^{2})}}\,dx={\frac {\pi }{4}}\log(1+a^{2})\qquad a\geq 0}$
ohne Beweis

##### 1.3
${\displaystyle \int _{0}^{\infty }{\frac {\arctan \alpha x}{x\,{\sqrt {1-x^{2}}}}}\,dx={\frac {\pi }{2}}\,{\text{arsinh}}\,\alpha }$
Beweis

In der Formel ${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {1}{a^{2}\,\cos ^{2}x+b^{2}\,\sin ^{2}x}}\,dx={\frac {\pi }{2ab}}}$ für ${\displaystyle a,b>0\,}$

setze ${\displaystyle a=1\,}$ und ${\displaystyle b={\sqrt {1+t^{2}}}}$.

${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {1}{\underbrace {\cos ^{2}x+\sin ^{2}x} _{=1}+t^{2}\,\sin ^{2}x}}\,dx={\frac {\pi }{2}}\,{\frac {1}{\sqrt {1+t^{2}}}}}$

Nach Substitution ${\displaystyle x\to \arcsin x\,}$ ist ${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {1}{1+t^{2}x^{2}}}\,{\frac {dx}{\sqrt {1-x^{2}}}}={\frac {\pi }{2}}\,{\frac {1}{\sqrt {1+t^{2}}}}}$.

Integriere nun nach ${\displaystyle t\,}$ von ${\displaystyle 0\,}$ bis ${\displaystyle \alpha \,:\,\,\int _{0}^{\frac {\pi }{2}}\left[{\frac {\arctan tx}{x}}\right]_{0}^{\alpha }\,{\frac {dx}{\sqrt {1-x^{2}}}}={\frac {\pi }{2}}\,{\text{arsinh}}\,\alpha }$.

##### 1.4
${\displaystyle \int _{0}^{\infty }{\frac {\arctan x}{1+2\cos \alpha \,\,x+x^{2}}}\,dx={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}\qquad -{\frac {\pi }{2}}<{\text{Re}}(\alpha )<{\frac {\pi }{2}}}$
Beweis

Nach Substitution ${\displaystyle x\mapsto {\frac {1}{x}}}$ lässt sich das Integral auch schreiben als ${\displaystyle \int _{0}^{\infty }{\frac {\arctan {\frac {1}{x}}}{1+2\cos \alpha \,\,x+x^{2}}}\,dx}$.

Addiert man beide Darstellungen, so ist ${\displaystyle 2I=\int _{0}^{\infty }{\frac {\arctan x+\arctan {\frac {1}{x}}}{1+2\cos \alpha \,\,x+x^{2}}}\,dx}$. Der Zähler ist konstant ${\displaystyle {\frac {\pi }{2}}}$.

Somit ist ${\displaystyle I={\frac {\pi }{4}}\int _{0}^{\infty }{\frac {1}{1+2\cos \alpha \,\,x+x^{2}}}\,dx={\frac {\pi }{4}}\left[{\frac {1}{\sin \alpha }}\arctan {\frac {x+\cos \alpha }{\sin \alpha }}\right]_{0}^{\infty }={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}}$.

##### 1.5
${\displaystyle \int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\arctan {\sqrt {2a^{2}+x^{2}}}}{\sqrt {2a^{2}+x^{2}}}}\,dx=\pi \,\arctan {\frac {1}{\sqrt {2a^{2}+1}}}-\left(\arctan {\frac {1}{a}}\right)^{2}}$
Beweis (Ahmedsches Integral)

Es ist ${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}={\frac {p+q}{p\,q}}\Rightarrow {\frac {1}{p\,(p+q)}}+{\frac {1}{q\,(p+q)}}={\frac {1}{p\,q}}}$.

Setze ${\displaystyle p=a^{2}+x^{2}\,}$ und ${\displaystyle q=a^{2}+y^{2}\,}$ und integriere nach ${\displaystyle x\,}$ und ${\displaystyle y\,}$ jeweils von ${\displaystyle 0\,}$ bis ${\displaystyle 1\,}$.

${\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {dx\,dy}{(a^{2}+x^{2})(2a^{2}+x^{2}+y^{2})}}+\int _{0}^{1}\int _{0}^{1}{\frac {dy\,dx}{(a^{2}+y^{2})(2a^{2}+x^{2}+y^{2})}}=\int _{0}^{1}{\frac {dx}{a^{2}+x^{2}}}\cdot \int _{0}^{1}{\frac {dy}{a^{2}+y^{2}}}}$

Vertauscht man die Rollen von ${\displaystyle x\,}$ und ${\displaystyle y\,}$, so erkennt man, dass beide Integrale auf der linken Seite gleich sind und dass beide Integrale auf der rechten Seite gleich sind.

Also ist ${\displaystyle 2\int _{0}^{1}\int _{0}^{1}{\frac {dx}{a^{2}+x^{2}}}\,{\frac {dy}{{\sqrt {2a^{2}+x^{2}}}^{2}+y^{2}}}=\left({\frac {1}{a}}\arctan {\frac {1}{a}}\right)^{2}}$

${\displaystyle \Rightarrow \int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,\left.{\frac {\arctan {\frac {y}{\sqrt {2a^{2}+x^{2}}}}}{\sqrt {2a^{2}+x^{2}}}}\right|_{0}^{1}\,dx=\left(\arctan {\frac {1}{a}}\right)^{2}}$.

Schreibe nun ${\displaystyle \arctan {\frac {1}{\sqrt {2a^{2}+x^{2}}}}}$ als ${\displaystyle {\frac {\pi }{2}}-\arctan {\sqrt {2a^{2}+x^{2}}}}$.

${\displaystyle \underbrace {\int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\frac {\pi }{2}}{\sqrt {2a^{2}+x^{2}}}}\,dx} _{\pi \left.\arctan {\frac {x}{\sqrt {2a^{2}+x^{2}}}}\right|_{0}^{1}}-\int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\arctan {\sqrt {2a^{2}+x^{2}}}}{\sqrt {2a^{2}+x^{2}}}}\,dx=\left(\arctan {\frac {1}{a}}\right)^{2}}$

##### 2.1
${\displaystyle \int _{0}^{\infty }{\frac {\arctan(x)}{\exp(\pi x)-1}}\,\mathrm {d} x={\frac {1}{2}}{\bigl [}1-\ln(2){\bigr ]}}$
ohne Beweis

##### 2.2
${\displaystyle \int _{0}^{\infty }{\frac {\arctan(x)}{\exp(2\pi x)-1}}\,\mathrm {d} x={\frac {1}{4}}{\bigl [}2-\ln(2\pi ){\bigr ]}}$
ohne Beweis

##### 2.3
${\displaystyle \int _{0}^{\infty }{\frac {\arctan(x)}{\sinh(\pi x)}}\,\mathrm {d} x={\frac {1}{2}}\ln {\bigl (}{\frac {\pi }{2}}{\bigr )}}$
ohne Beweis

##### 2.4
${\displaystyle \int _{0}^{\infty }{\frac {2\arctan(x)-x\ln(x^{2}+1)}{2\,(x^{2}+1)\sinh(\pi x)}}\,\mathrm {d} x={\frac {1}{2}}{\bigl [}2\,\gamma \ln(2)-\ln(2)^{2}{\bigr ]}}$
ohne Beweis