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

Zurück zu Bestimmte Integrale

0.1

${\int }_0^1\frac{\text{artanh}x\log x}{x(1-x)(1+x)}dx=-\frac{1}{16}(7\zeta (3)+2{\pi }^2\log 2)$

Beweis

$\text{artanh}x=\frac{1}{2}\cdot \log \left(\frac{1+x}{1-x}\right)=\frac{1}{2}\cdot [\log (1+x)-\log (1-x)]$

$\Rightarrow \text{artanh}x\cdot \log x=\frac{1}{2}\cdot \log (1+x)\log x-\frac{1}{2}\cdot \log (1-x)\log x$

$\frac{1}{x(1-x)(1+x)}=\frac{1}{x}+\frac{1}{2}\cdot \frac{1}{1-x}-\frac{1}{2}\cdot \frac{1}{1+x}$


$\begin{array}{cc}\frac{\text{artanh}x\cdot \log x}{x(1-x)(1+x)}=& +\frac{1}{2}\cdot \frac{\log (1+x)\log x}{x}+\frac{1}{4}\cdot \frac{\log (1+x)\log x}{1-x}-\frac{1}{4}\cdot \frac{\log (1+x)\log x}{1+x}\\ \\ & -\frac{1}{2}\cdot \frac{\log (1-x)\log x}{x}-\frac{1}{4}\cdot \frac{\log (1-x)\log x}{1-x}+\frac{1}{4}\cdot \frac{\log (1-x)\log x}{1+x}\end{array}$



$\begin{array}{cc}{\int }_0^1\frac{\text{artanh}x\cdot \log x}{x(1-x)(1+x)}dx=& +\frac{1}{2}\cdot \underset{=-\frac{3}{4}\zeta (3)}{\underbrace{{\int }_0^1\frac{\log (1+x)\log x}{x}dx}}+\frac{1}{4}\cdot \underset{=-\frac{{\pi }^2}{4}\log 2+\zeta (3)}{\underbrace{{\int }_0^1\frac{\log (1+x)\log x}{1-x}dx}}-\frac{1}{4}\cdot \underset{-\frac{1}{8}\zeta (3)}{\underbrace{{\int }_0^1\frac{\log (1+x)\log x}{1+x}dx}}\\ \\ & -\frac{1}{2}\cdot \underset{=\zeta (3)}{\underbrace{{\int }_0^1\frac{\log (1-x)\log x}{x}dx}}-\frac{1}{4}\cdot \underset{=\zeta (3)}{\underbrace{{\int }_0^1\frac{\log (1-x)\log x}{1-x}dx}}+\frac{1}{4}\cdot \underset{=-\frac{{\pi }^2}{4}\log 2+\frac{13}{8}\zeta (3)}{\underbrace{{\int }_0^1\frac{\log (1-x)\log x}{1+x}dx}}\end{array}$


${\int }_0^1\frac{\text{artanh}x\cdot \log x}{x(1-x)(1+x)}dx=-\frac{3}{8}\zeta (3)-\frac{{\pi }^2}{16}\log 2+\frac{1}{4}\zeta (3)+\frac{1}{32}\zeta (3)-\frac{1}{2}\zeta (3)-\frac{1}{4}\zeta (3)-\frac{{\pi }^2}{16}\log 2+\frac{13}{32}\zeta (3)=-\frac{7}{16}\zeta (3)-\frac{{\pi }^2}{8}\log 2$