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

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0.1

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

Beweis

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

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

0.2

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

Beweis

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

0.3

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

ohne Beweis

0.4

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

ohne Beweis

0.5

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

ohne Beweis

0.6

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

ohne Beweis

1.1

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

Beweis

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

1.2

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

ohne Beweis

1.3

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

Beweis

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

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

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

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

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

1.4

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

Beweis

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

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

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

1.5

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

Beweis (Ahmedsches Integral)

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

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

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

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

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

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

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

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

2.1

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

ohne Beweis

2.2

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

ohne Beweis

2.3

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

ohne Beweis

2.4

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

ohne Beweis