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

Zurück zu Bestimmte Integrale

##### 0.1
${\displaystyle \int _{0}^{1}{\frac {\log(1+x)-\log 2}{1+x^{2}}}\,dx=-{\frac {\pi }{8}}\log 2}$
ohne Beweis

##### 0.2
${\displaystyle \int _{0}^{1}{\frac {\log(1+x)-\log 2}{1-x^{2}}}\,dx=-{\frac {\pi ^{2}}{24}}}$
ohne Beweis

##### 0.3
${\displaystyle \int _{1}^{\infty }{\frac {\log(1+x)-\log 2}{1+x^{2}}}\,dx=G-{\frac {\pi }{8}}\log 2}$
ohne Beweis

##### 0.4
${\displaystyle \int _{1}^{\infty }{\frac {\log(1+x)-\log 2}{1-x^{2}}}\,dx=-{\frac {\pi ^{2}}{12}}}$
ohne Beweis

##### 0.5
${\displaystyle \int _{0}^{1}{\frac {\log x}{1+x^{2}}}\,dx=-G}$
Beweis

Aus ${\displaystyle {\frac {\log x}{1+x^{2}}}=\sum _{n=0}^{\infty }(-1)^{n}\,x^{2n}\,\log x}$

folgt ${\displaystyle \int _{0}^{1}{\frac {\log x}{1+x^{2}}}dx=\sum _{n=0}^{\infty }(-1)^{n}\int _{0}^{1}x^{2n}\log x\,dx=\sum _{n=0}^{\infty }(-1)^{n}\,{\frac {-1}{(2n+1)^{2}}}=-G}$.

##### 0.6
${\displaystyle \int _{0}^{1}{\frac {\log x}{1-x^{2}}}\,dx=-{\frac {\pi ^{2}}{8}}}$
Beweis

Aus ${\displaystyle {\frac {\log x}{1-x^{2}}}=\sum _{n=0}^{\infty }x^{2n}\,\log x}$

folgt ${\displaystyle \int _{0}^{1}{\frac {\log x}{1-x^{2}}}dx=\sum _{n=0}^{\infty }\int _{0}^{1}x^{2n}\log x\,dx=\sum _{n=0}^{\infty }{\frac {-1}{(2n+1)^{2}}}=-{\frac {\pi ^{2}}{8}}}$.

##### 0.7
${\displaystyle \int _{0}^{\infty }{\frac {\log(1+x+x^{2})}{1+x^{2}}}\,dx={\frac {\pi }{3}}\log(2+{\sqrt {3}})+{\frac {4}{3}}\,G}$
Beweis

In der Formel ${\displaystyle \int _{0}^{\infty }{\frac {\log(1+2\sin \alpha \cdot x+x^{2})}{1+x^{2}}}\,dx=\pi \log \left(2\cos {\frac {\alpha }{2}}\right)+\alpha \log \left(\tan {\frac {\alpha }{2}}\right)+2\sum _{k=0}^{\infty }{\frac {\sin(2k+1)\alpha }{(2k+1)^{2}}}}$ setze ${\displaystyle \alpha ={\frac {\pi }{6}}}$.

Wegen ${\displaystyle \pi \log \left(2\cos {\frac {\pi }{12}}\right)={\frac {\pi }{2}}\left(\left(2\cos {\frac {\pi }{12}}\right)^{2}\right)={\frac {\pi }{2}}\log(2+{\sqrt {3}})\,,\;{\frac {\pi }{6}}\log \left(\tan {\frac {\pi }{12}}\right)=-{\frac {\pi }{6}}\log \left(\cot {\frac {\pi }{12}}\right)=-{\frac {\pi }{6}}\log(2+{\sqrt {3}})}$

und ${\displaystyle 2\sum _{k=0}^{3N}{\frac {\sin(2k+1){\frac {\pi }{6}}}{(2k+1)^{2}}}=\sum _{k=0}^{3N}{\frac {(-1)^{k}}{(2k+1)^{2}}}+3\sum _{k=0}^{N-1}{\frac {(-1)^{k}}{(6k+3)^{2}}}\to G+{\frac {G}{3}}={\frac {4}{3}}\,G}$

ist dann ${\displaystyle \int _{0}^{\infty }{\frac {\log(1+x+x^{2})}{1+x^{2}}}\,dx={\frac {\pi }{3}}\log(2+{\sqrt {3}})+{\frac {4}{3}}\,G}$.

##### 0.8
${\displaystyle \int _{0}^{1}\log(-\log x)\,dx=-\gamma }$
Beweis

Differenziere ${\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}\,e^{-t}\,dt}$.

${\displaystyle \Gamma '(z)=\int _{0}^{\infty }t^{z-1}\,\log(t)\,e^{-t}\,dt}$.

und setze ${\displaystyle z=1\,}$.

${\displaystyle -\gamma =\Gamma '(1)=\int _{0}^{\infty }\log(t)\,e^{-t}\,dt}$.

Dies ist nach Substitution ${\displaystyle t\mapsto -\log x}$ gleich ${\displaystyle \int _{0}^{1}\log(-\log x)\,dx}$.

##### 0.9
${\displaystyle \int _{0}^{1}\left({\frac {2\log x}{x^{2}-4x+8}}-{\frac {3\log x}{x^{2}+2x+2}}\right)dx=G}$
ohne Beweis

##### 0.10
${\displaystyle \int _{0}^{\infty }{\frac {\log(x+1)}{\log ^{2}x+\pi ^{2}}}\,{\frac {dx}{x^{2}}}=\gamma }$
Beweis

Betrachte die Formel ${\displaystyle \int _{0}^{\infty }{\frac {1}{x+1-u}}\cdot {\frac {1}{\log ^{2}x+\pi ^{2}}}\,dx={\frac {1}{u}}+{\frac {1}{\log(1-u)}}}$.

Wegen ${\displaystyle \int _{0}^{1}{\frac {1}{x+1-u}}\,du=\left[-\log(x+1-u)\right]_{0}^{1}=\log \left(1+{\frac {1}{x}}\right)}$

und ${\displaystyle \int _{0}^{1}\left[{\frac {1}{u}}+{\frac {1}{\log(1-u)}}\right]du=\gamma }$ ist ${\displaystyle \int _{0}^{\infty }{\frac {\log \left(1+{\frac {1}{x}}\right)}{\log ^{2}x+\pi ^{2}}}\,dx=\gamma }$.

Nach der Substitution ${\displaystyle x\mapsto {\frac {1}{x}}}$ erhält man das gesuchte Integral.

##### 0.11
${\displaystyle \int _{0}^{1}{\frac {\log \left(1+x^{2+{\sqrt {3}}}\,\right)}{1+x}}\,dx={\frac {\pi ^{2}}{12}}\cdot \left(1-{\sqrt {3}}\,\right)+\log(2)\cdot \log \left(1+{\sqrt {3}}\,\right)}$
ohne Beweis

##### 0.12
${\displaystyle \int _{0}^{1}{\frac {\log \left(1+x^{4+{\sqrt {15}}}\,\right)}{1+x}}\,dx={\frac {\pi ^{2}}{12}}\cdot \left(2-{\sqrt {15}}\,\right)+\log \left({\frac {1+{\sqrt {5}}}{2}}\right)\cdot \log \left(2+{\sqrt {3}}\,\right)+\log(2)\cdot \log \left({\sqrt {3}}+{\sqrt {5}}\,\right)}$
ohne Beweis

##### 0.13
${\displaystyle \int _{0}^{1}{\frac {\log \left(1+x^{6+{\sqrt {35}}}\,\right)}{1+x}}\,dx={\frac {\pi ^{2}}{12}}\cdot \left(3-{\sqrt {35}}\,\right)+\log \left({\frac {1+{\sqrt {5}}}{2}}\right)\cdot \log \left(8+3{\sqrt {7}}\,\right)+\log(2)\cdot \log \left({\sqrt {5}}+{\sqrt {7}}\,\right)}$
ohne Beweis

##### 0.14
${\displaystyle \int _{0}^{1}\log(1-x)\,\log(x)\,\log(1+x)\,dx=-6+{\frac {5\pi ^{2}}{12}}-(\log 2)^{2}+4\cdot \log 2-{\frac {\pi ^{2}}{2}}\log 2+{\frac {21}{8}}\zeta (3)}$
ohne Beweis

##### 0.15
${\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{1+x}}\,dx=-{\frac {\zeta (3)}{8}}}$
Beweis

Wegen ${\displaystyle {\frac {\log(1+x)}{1+x}}=\sum _{k=1}^{\infty }(-1)^{k-1}\,H_{k}\ x^{k}}$ ist

${\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{1+x}}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,H_{k}\,\int _{0}^{1}x^{k}\,\log x\,\,dx=\sum _{k=1}^{\infty }(-1)^{k}\,{\frac {H_{k}}{(k+1)^{2}}}=-{\frac {\zeta (3)}{8}}}$.

##### 0.16
${\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{x}}\,dx=-{\frac {3}{4}}\,\zeta (3)}$
Beweis

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

${\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{x}}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,{\frac {1}{k}}\,\int _{0}^{1}x^{k-1}\,\log x\,dx=\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k^{3}}}=-{\frac {3}{4}}\,\zeta (3)}$

##### 0.17
${\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{1-x}}\,dx=-{\frac {\pi ^{2}}{4}}\log 2+\zeta (3)}$
Beweis

${\displaystyle I:=\int _{0}^{1}{\frac {\log(1+x)\,\log x}{1-x}}\,dx=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}\int _{0}^{1}{\frac {x^{n}\,\log x}{1-x}}\,dx}$,

wobei ${\displaystyle \int _{0}^{1}{\frac {x^{n}\,\log x}{1-x}}\,dx=\sum _{k=0}^{\infty }\int _{0}^{1}x^{n+k}\,\log x\,dx=\sum _{k=0}^{\infty }{\frac {-1}{(n+k+1)^{2}}}=\sum _{k=1}^{n}{\frac {1}{k^{2}}}-{\frac {\pi ^{2}}{6}}}$ ist.

${\displaystyle I=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}\left(H_{n,2}-{\frac {\pi ^{2}}{6}}\right)=\underbrace {\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}\,H_{n,2}} _{=-{\frac {\pi ^{2}}{12}}\log 2+\zeta (3)}-{\frac {\pi ^{2}}{6}}\log 2=-{\frac {\pi ^{2}}{4}}\log 2+\zeta (3)}$

##### 0.18
${\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log x}{1+x}}\,dx=-{\frac {\pi ^{2}}{4}}\log 2+{\frac {13}{8}}\,\zeta (3)}$
ohne Beweis

##### 0.19
${\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log x}{x}}\,dx=\zeta (3)}$
ohne Beweis

##### 0.20
${\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log x}{1-x}}\,dx=\zeta (3)}$
ohne Beweis

##### 0.21
${\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log(1+x)}{x}}\,dx=-{\frac {5}{8}}\,\zeta (3)}$
ohne Beweis

##### 0.22
${\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log(1+x)}{1+x}}\,dx={\frac {(\log 2)^{2}}{8}}-{\frac {\pi ^{2}}{12}}\log 2+{\frac {\zeta (3)}{8}}}$
ohne Beweis

##### 1.1
${\displaystyle \int _{0}^{1}{\frac {\log ^{2n}x}{1+x^{2}}}\,dx={\frac {1}{2}}\,|E_{2n}|\,\left({\frac {\pi }{2}}\right)^{2n+1}\qquad n\in \mathbb {Z} ^{\geq 0}}$
ohne Beweis

##### 1.2
${\displaystyle \int _{0}^{\infty }{\frac {\log ^{n-1}x}{1+x^{2}}}\,dx=|E_{n-1}|\,\left({\frac {\pi }{2}}\right)^{n}\qquad n\in \mathbb {Z} ^{\geq 1}}$
Beweis

In der Formel ${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx={\frac {\pi }{\beta }}\csc \left({\frac {\alpha \pi }{\beta }}\right)}$ setze ${\displaystyle \beta =2\,}$ und verschiebe ${\displaystyle \alpha \,}$ um ${\displaystyle 1\,}$ nach rechts.

${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha }}{1+x^{2}}}\,dx={\frac {\pi }{2}}\,\csc \left({\frac {(\alpha +1)\pi }{2}}\right)={\frac {\pi }{2}}\sec \left({\frac {\alpha \pi }{2}}\right)={\frac {\pi }{2}}\sum _{k=0}^{\infty }|E_{k}|\,{\frac {1}{k!}}\,\left({\frac {\alpha \pi }{2}}\right)^{k}=\sum _{k=0}^{\infty }|E_{k}|\,\left({\frac {\pi }{2}}\right)^{k+1}\,{\frac {\alpha ^{k}}{k!}}}$

Differenziere ${\displaystyle n-1\,}$ mal nach ${\displaystyle \alpha \,}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha }\,\log ^{n-1}x}{1+x^{2}}}\,dx=\sum _{k=n-1}^{\infty }|E_{k}|\,\left({\frac {\pi }{2}}\right)^{k+1}\,{\frac {\alpha ^{k-n+1}}{(k-n+1)!}}}$

Und setze ${\displaystyle \alpha =0\,}$

${\displaystyle \int _{0}^{\infty }{\frac {\log ^{n-1}x}{1+x^{2}}}\,dx=|E_{n-1}|\,\left({\frac {\pi }{2}}\right)^{n}}$

##### 1.3
${\displaystyle \int _{0}^{\infty }{\frac {\log ^{n-1}x}{1-x^{2}}}\,dx={\frac {2^{n}(1-2^{n})|B_{n}|}{n}}\,\left({\frac {\pi }{2}}\right)^{n}\qquad n\in \mathbb {Z} ^{\geq 1}}$
ohne Beweis

##### 1.4
${\displaystyle \int _{0}^{\infty }{\frac {dx}{(x+1)^{n}\,(\log ^{2}x+\pi ^{2})}}=C_{n}}$ Fontana-Zahlen genügen der Rekursion: ${\displaystyle \quad C_{0}=-1,\quad \sum _{k=0}^{n-1}{\frac {C_{k}}{n-k}}=0}$
ohne Beweis

##### 1.5
${\displaystyle \int _{0}^{\infty }{\frac {1}{x+1-u}}\cdot {\frac {1}{\log ^{2}x+\pi ^{2}}}\,dx={\frac {1}{u}}+{\frac {1}{\log(1-u)}}\qquad u\in \mathbb {C} ^{\times }\setminus \mathbb {R} ^{\geq 1}}$
Beweis

Die Funktion ${\displaystyle f(z)={\frac {1}{z+1-u}}\cdot {\frac {1}{\log(-z)}}}$ ist auf ${\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}$ meromorph.

Die Polstellen liegen bei ${\displaystyle -1\,}$ und ${\displaystyle u-1\,}$. Dabei ist ${\displaystyle {\text{res}}(f,-1)={\frac {1}{u}}}$ und ${\displaystyle {\text{res}}(f,u-1)={\frac {1}{\log(1-u)}}}$.

Also gilt nach dem Residuensatz ${\displaystyle \int _{\gamma _{R,\varepsilon }}f\,dz+\int _{C_{R}}f\,dz+\int _{\delta _{R,\varepsilon }}f\,dz+\int _{c_{\varepsilon }}f\,dz=2\pi i\,\left({\frac {1}{u}}+{\frac {1}{\log(1-u)}}\right)}$.

Aus ${\displaystyle L(C_{R})\sim 2\pi R\,}$ und ${\displaystyle M(C_{R})=\max _{z\in C_{R}}|f(z)|\sim {\frac {1}{R\,\log R}}}$ folgt ${\displaystyle \left|\int _{C_{R}}f\,dz\right|\leq L(C_{R})\,M(C_{R})\sim {\frac {2\pi }{\log R}}}$.

Daher geht ${\displaystyle \int _{C_{R}}fdz}$ gegen null für ${\displaystyle R\to \infty \,}$.

Für ${\displaystyle z\,}$, nahe bei 0, ist ${\displaystyle \log(-z)\,}$ groß, und somit ${\displaystyle f(z)={\frac {1}{z+1-u}}\cdot {\frac {1}{\log(-z)}}}$ klein.

Daher geht ${\displaystyle \int _{c_{\varepsilon }}f\,dz}$ für ${\displaystyle \varepsilon \to 0+\,}$ auch gegen null.

Im Grenzübergang ${\displaystyle R\to \infty ,\varepsilon \to 0+\,}$ ergibt sich:

${\displaystyle \int _{0}^{\infty }{\frac {1}{x+1-u}}\left({\frac {1}{\log(-x-i0^{+})}}-{\frac {1}{\log(-x+i0^{+})}}\right)dx=2\pi i\left({\frac {1}{u}}+{\frac {1}{\log(1-u)}}\right)}$.

Dabei ist ${\displaystyle {\frac {1}{\log x-i\pi }}-{\frac {1}{\log x+i\pi }}={\frac {2\pi i}{\log ^{2}x+\pi ^{2}}}}$, und somit gilt

${\displaystyle \int _{0}^{\infty }{\frac {1}{x+1-u}}\cdot {\frac {1}{\log ^{2}x+\pi ^{2}}}\,dx={\frac {1}{u}}+{\frac {1}{\log(1-u)}}}$.

##### 1.6
${\displaystyle \int _{0}^{1}{\frac {\left(\log \,{\frac {1}{x}}\right)^{z-1}}{1+x}}\,dx=\eta (z)\,\Gamma (z)\qquad {\text{Re}}(z)>0}$
ohne Beweis

##### 1.7
${\displaystyle \int _{0}^{1}{\frac {\left(\log \,{\frac {1}{x}}\right)^{z-1}}{1-x}}\,dx=\zeta (z)\,\Gamma (z)\qquad {\text{Re}}(z)>1}$
ohne Beweis

##### 1.8
${\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}-x^{-\alpha }}{(x+1)\,\log x}}\,dx=\log \left(\tan {\frac {\alpha \pi }{2}}\right)\qquad 0<\mathrm {Re} (\alpha )<1}$
Beweis

${\displaystyle \int _{0}^{1}{\frac {x^{-\alpha }}{1+x}}\,dx}$ ist nach Substitution ${\displaystyle x\mapsto {\frac {1}{x}}}$ gleich ${\displaystyle \int _{1}^{\infty }{\frac {x^{\alpha -1}}{1+x}}\,dx}$.

Also ist ${\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}+x^{-\alpha }}{1+x}}\,dx=\int _{0}^{1}{\frac {x^{\alpha -1}}{1+x}}\,dx+\int _{1}^{\infty }{\frac {x^{\alpha -1}}{1+x}}\,dx}$   ${\displaystyle =\int _{0}^{\infty }{\frac {x^{\alpha -1}}{1+x}}\,dx={\frac {\pi }{\sin \alpha \pi }}}$.

Integriere nach ${\displaystyle \alpha \,}$:

${\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}-x^{-\alpha }}{(1+x)\log x}}\,dx=\log \left(\tan {\frac {\alpha \pi }{2}}\right)+C}$

Dass dabei ${\displaystyle C=0\,}$ ist, erkennt man, wenn man ${\displaystyle \alpha ={\frac {1}{2}}}$ setzt.

##### 1.9
${\displaystyle \int _{0}^{1}\left(\log {\frac {1}{x}}\right)^{z-1}dx=\Gamma (z)\qquad {\text{Re}}(z)>0}$
Beweis

In der Formel ${\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt}$ substituiere ${\displaystyle t=-\log x\,}$:

${\displaystyle \Gamma (z)=\int _{0}^{1}\left(-\log x\right)^{z-1}dx}$

##### 1.10
${\displaystyle \int _{0}^{1}\left({\frac {\log x}{a+1-x}}-{\frac {\log x}{a+x}}\right)dx={\frac {1}{2}}\left(\log a-\log(a+1)\right)^{2}\qquad \forall a\in \mathbb {C} \setminus [-1,0]}$
Beweis

Definiert man ${\displaystyle F:\,]0,1]\to \mathbb {C} }$ als ${\displaystyle F(x)={\frac {x\log x}{a+x}}-\log(a+x)}$,

so ist ${\displaystyle F'(x)={\frac {\log x+1}{a+x}}-{\frac {x\log x}{(a+x)^{2}}}-{\frac {1}{a+x}}}$

${\displaystyle ={\frac {a\log x+a+x\log x+x-x\log x-(a+x)}{(a+x)^{2}}}={\frac {a\log x}{(a+x)^{2}}}}$.

Also ist ${\displaystyle a\int _{0}^{1}{\frac {\log x}{(a+x)^{2}}}\,dx=F(1)-F(0+)=\log a-\log(a+1)}$.

Definiert man ${\displaystyle G:\,]0,1]\to \mathbb {C} }$ als ${\displaystyle G(x)={\frac {x\log x}{a+1-x}}+\log(a+1-x)}$,

so ist ${\displaystyle G'(x)={\frac {\log x+1}{a+1-x}}+{\frac {x\log x}{(a+1-x)^{2}}}-{\frac {1}{a+1-x}}}$

${\displaystyle ={\frac {(a+1)\log x+a+1-x\log x-x+x\log x-(a+1-x)}{(a+1-x)^{2}}}={\frac {(a+1)\log x}{(a+1-x)^{2}}}}$

Also ist ${\displaystyle (a+1)\int _{0}^{1}{\frac {\log x}{(a+1-x)^{2}}}\,dx=G(1)-G(0+)=\log a-\log(a+1)}$.

Mit den beiden Integralen erhält man ${\displaystyle \int _{0}^{1}\left({\frac {\log x}{(a+x)^{2}}}-{\frac {\log x}{(a+1-x)^{2}}}\right)dx=\left(\log a-\log(a+1)\right)\left({\frac {1}{a}}-{\frac {1}{a+1}}\right)}$.

Integriert man beide Seiten nach ${\displaystyle a\,}$, so ist ${\displaystyle \int _{0}^{1}\left(-{\frac {\log x}{a+x}}+{\frac {\log x}{a+1-x}}\right)dx={\frac {1}{2}}\left(\log a-\log(a+1)\right)^{2}+C}$.

Dass die Konstante ${\displaystyle C=0\,}$ sein muss, erkennt man wenn man ${\displaystyle a\to \infty \,}$ gehen lässt.

##### 1.11
${\displaystyle \int _{0}^{1}{\frac {2\log x}{(a\pi )^{2}+\log ^{2}x}}\,{\frac {x}{1-x^{2}}}\,dx={\frac {1}{2a}}+\psi (a)-\log a\qquad {\text{Re}}(a)>0}$
Beweis

${\displaystyle \int _{0}^{1}{\frac {2\log x}{(a\pi )^{2}+\log ^{2}x}}\,{\frac {x}{1-x^{2}}}\,dx=\int _{0}^{1}}$   ${\displaystyle \int _{0}^{\infty }2\sin(t\log x)\,e^{-a\pi t}\,dt\,}$   ${\displaystyle \,{\frac {x}{1-x^{2}}}\,dx}$

${\displaystyle =\int _{0}^{\infty }\int _{0}^{1}2\sin(t\log x)\,{\frac {x}{1-x^{2}}}\,dx\,e^{-a\pi t}\,dt\qquad {\text{//subst:}}\,\,x=e^{-u}}$

${\displaystyle =\int _{0}^{\infty }\int _{0}^{\infty }{\frac {2\sin(tu)}{1-e^{2u}}}\,du\,e^{-a\pi t}\,dt=\int _{0}^{\infty }}$   ${\displaystyle \left({\frac {1}{t}}-{\frac {\pi }{2}}\coth {\frac {\pi t}{2}}\right)}$   ${\displaystyle e^{-a\pi t}\,dt\qquad {\text{//subst:}}\,\,\omega =\pi t}$

${\displaystyle =\int _{0}^{\infty }\left({\frac {1}{\omega }}-{\frac {1}{e^{\omega }-1}}-{\frac {1}{2}}\right)e^{-a\omega }\,d\omega =}$   ${\displaystyle \psi (a+1)-\log a-{\frac {1}{2a}}}$

##### 1.12
${\displaystyle \int _{0}^{\infty }{\frac {\log(1+2\sin \alpha \,\,x+x^{2})}{1+x^{2}}}\,dx=\pi \log \left(2\cos {\frac {\alpha }{2}}\right)+\alpha \log \left(\tan {\frac {\alpha }{2}}\right)+2\sum _{k=0}^{\infty }{\frac {\sin(2k+1)\alpha }{(2k+1)^{2}}}\qquad 0<\alpha <\pi }$
Beweis

Setzt man ${\displaystyle F(z):=\int _{0}^{\infty }{\frac {\log(1+2\sin z\;\,x+x^{2})}{1+x^{2}}}\,dx}$,

so ist ${\displaystyle F(0)=\int _{0}^{\infty }{\frac {\log(1+x^{2})}{1+x^{2}}}\,dx=\int _{0}^{\frac {\pi }{2}}\log(1+\tan ^{2}x)\,dx=-2\int _{0}^{\frac {\pi }{2}}\log(\cos x)\,dx=\pi \log 2}$

und ${\displaystyle F'(z)=\int _{0}^{\infty }{\frac {1}{1+x^{2}}}\,{\frac {2x\cos z}{1+2\sin z\;\,x+x^{2}}}\,dx={\frac {\cos z}{\sin z}}\int _{0}^{\infty }{\frac {1}{1+x^{2}}}\,{\frac {2x\sin z}{1+2\sin z\;\,x+x^{2}}}\,dx}$

${\displaystyle ={\frac {\cos z}{\sin z}}\int _{0}^{\infty }\left({\frac {1}{1+x^{2}}}-{\frac {1}{1+2\sin z\;\,x+x^{2}}}\right)dx={\frac {\pi }{2}}\cdot {\frac {\cos z}{\sin z}}-{\frac {\cos z}{\sin z}}\left[{\frac {1}{\cos z}}\,\arctan {\frac {x+\sin z}{\cos z}}\right]_{0}^{\infty }}$

${\displaystyle ={\frac {\pi }{2}}\cdot {\frac {\cos z}{\sin z}}-{\frac {\pi }{2}}\cdot {\frac {1}{\sin z}}+{\frac {z}{\sin z}}=-{\frac {\pi }{2}}\tan {\frac {z}{2}}+{\frac {z}{\sin z}}}$.

Nun ist ${\displaystyle F(\alpha )-\pi \log 2=\int _{0}^{\alpha }F'(z)\,dz=\pi \log \left(\cos {\frac {\alpha }{2}}\right)+\int _{0}^{\alpha }{\frac {z}{\sin z}}\,dz}$

und somit ist ${\displaystyle F(\alpha )-\pi \log \left(2\cos {\frac {\alpha }{2}}\right)=\int _{0}^{\alpha }{\frac {z}{\sin z}}\,dz=\left[z\log \left(\tan {\frac {z}{2}}\right)\right]_{0}^{\alpha }-\int _{0}^{\alpha }\log \left(\tan {\frac {z}{2}}\right)dz}$

${\displaystyle =\alpha \log \left(\tan {\frac {\alpha }{2}}\right)+2\int _{0}^{\alpha }\sum _{k=0}^{\infty }{\frac {\cos(2k+1)z}{2k+1}}\,dz}$.

Daraus folgt ${\displaystyle F(\alpha )=\pi \log \left(2\cos {\frac {\alpha }{2}}\right)+\alpha \log \left(\tan {\frac {\alpha }{2}}\right)+2\sum _{k=0}^{\infty }{\frac {\sin(2k+1)\alpha }{(2k+1)^{2}}}}$.

##### 1.13
${\displaystyle \int _{0}^{1}{\frac {\log \log \left({\frac {1}{x}}\right)}{1+2\cos \alpha \pi \cdot x+x^{2}}}\,dx={\frac {\pi }{2\sin \alpha \pi }}\left(\alpha \log 2\pi +\log {\frac {\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{2}}-{\frac {\alpha }{2}}\right)}}\right)\qquad 0<\alpha <1}$
Beweis

${\displaystyle {\frac {\sin \alpha \pi }{1+2\cos \alpha \pi \cdot x+x^{2}}}=\sum _{n=1}^{\infty }(-x)^{n-1}\,\sin n\pi \alpha \qquad \quad |x|<1}$

${\displaystyle \int _{0}^{1}{\frac {\sin \alpha \pi }{1+2\cos \alpha \pi \cdot x+x^{2}}}\,\log \log \left({\frac {1}{x}}\right)dx=\sum _{n=1}^{\infty }(-1)^{n-1}\int _{0}^{1}x^{n-1}\,\log \log \left({\frac {1}{x}}\right)dx\,\;\sin n\pi \alpha }$

${\displaystyle =\sum _{n=1}^{\infty }(-1)^{n}\,{\frac {\gamma +\log n}{n}}\,\sin n\pi \alpha =\gamma \cdot \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}\,\sin n\pi \alpha +\sum _{n=1}^{\infty }(-1)^{n}\,{\frac {\log n}{n}}\,\sin n\pi \alpha }$

${\displaystyle -{\frac {\alpha \pi }{2}}\,\gamma +{\frac {\alpha \pi }{2}}\left(\gamma +\log 2\pi \right)+{\frac {\pi }{2}}\log {\frac {\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{2}}-{\frac {\alpha }{2}}\right)}}={\frac {\pi }{2}}\left(\alpha \log 2\pi +\log {\frac {\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{2}}-{\frac {\alpha }{2}}\right)}}\right)}$

##### 1.14
${\displaystyle \int _{0}^{1}{\frac {\log(1-x)}{x}}\,{\frac {2z}{\log ^{2}x+(2\pi z)^{2}}}\,dx=-\log \left({\frac {z!\,e^{z}}{z^{z}\,{\sqrt {2\pi z}}}}\right)\qquad {\text{Re}}(z)>0}$
Beweis

##### 2.1
${\displaystyle \int _{a}^{b}{\frac {\log x}{(x+a)(x+b)}}\,dx={\frac {\log(ab)}{2\,(b-a)}}\,\log \left({\frac {(a+b)^{2}}{4ab}}\right)}$
Beweis

Nach der Substitution ${\displaystyle x\mapsto {\frac {ab}{x}}}$ wird das Integral zu ${\displaystyle I=\int _{a}^{b}{\frac {\log(ab)-\log x}{(b+x)(a+x)}}\,dx}$

Also ist ${\displaystyle 2I=\int _{a}^{b}{\frac {\log(ab)}{(x+a)(x+b)}}\,dx={\frac {\log(ab)}{b-a}}\int _{a}^{b}\left({\frac {1}{x+a}}-{\frac {1}{x+b}}\right)\,dx}$

${\displaystyle ={\frac {\log(ab)}{b-a}}\,{\Big [}\log(x+a)-\log(x+b){\Big ]}_{a}^{b}={\frac {\log(ab)}{b-a}}\log \left({\frac {(a+b)^{2}}{4ab}}\right)}$.

##### 2.2
${\displaystyle \int _{0}^{\infty }{\frac {\log x}{(x+a)(x+b)}}\,dx={\frac {\log ^{2}(a)-\log ^{2}(b)}{2\,(a-b)}}}$
Beweis

Nach der Substitution ${\displaystyle x\mapsto {\frac {ab}{x}}}$ wird das Integral zu ${\displaystyle I=\int _{0}^{\infty }{\frac {\log(ab)-\log x}{(b+x)(a+x)}}\,dx}$.

Also ist ${\displaystyle 2I=\int _{0}^{\infty }{\frac {\log(ab)}{(x+a)(x+b)}}\,dx={\frac {\log(ab)}{a-b}}\int _{0}^{\infty }\left({\frac {1}{x+b}}-{\frac {1}{x+a}}\right)\,dx}$

${\displaystyle ={\frac {\log(a)+\log(b)}{a-b}}\,\underbrace {{\Big [}\log(x+b)-\log(x+a){\Big ]}_{0}^{\infty }} _{\log(a)-\log(b)}={\frac {\log ^{2}(a)-\log ^{2}(b)}{a-b}}}$.