Zurück zu Bestimmte Integrale
artanhx=12⋅log(1+x1−x)=12⋅[log(1+x)−log(1−x)]{\displaystyle {\text{artanh}}\,x={\frac {1}{2}}\cdot \log \left({\frac {1+x}{1-x}}\right)={\frac {1}{2}}\cdot {\Big [}\log(1+x)-\log(1-x){\Big ]}} ⇒artanhx⋅logx=12⋅log(1+x)logx−12⋅log(1−x)logx{\displaystyle \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} 1x(1−x)(1+x)=1x+12⋅11−x−12⋅11+x{\displaystyle {\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}}} artanhx⋅logxx(1−x)(1+x)=+12⋅log(1+x)logxx+14⋅log(1+x)logx1−x−14⋅log(1+x)logx1+x−12⋅log(1−x)logxx−14⋅log(1−x)logx1−x+14⋅log(1−x)logx1+x{\displaystyle {\begin{aligned}{\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{aligned}}} ∫01artanhx⋅logxx(1−x)(1+x)dx=+12⋅∫01log(1+x)logxxdx⏟=−34ζ(3)+14⋅∫01log(1+x)logx1−xdx⏟=−π24log2+ζ(3)−14⋅∫01log(1+x)logx1+xdx⏟−18ζ(3)−12⋅∫01log(1−x)logxxdx⏟=ζ(3)−14⋅∫01log(1−x)logx1−xdx⏟=ζ(3)+14⋅∫01log(1−x)logx1+xdx⏟=−π24log2+138ζ(3){\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {{\text{artanh}}\,x\cdot \log x}{x\,(1-x)\,(1+x)}}\,dx=&+{\frac {1}{2}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{x}}\,dx} _{=-{\frac {3}{4}}\zeta (3)}+{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{1-x}}\,dx} _{=-{\frac {\pi ^{2}}{4}}\log 2+\zeta (3)}-{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{1+x}}\,dx} _{-{\frac {1}{8}}\zeta (3)}\\\\&-{\frac {1}{2}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{x}}\,dx} _{=\zeta (3)}-{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{1-x}}\,dx} _{=\zeta (3)}+{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{1+x}}\,dx} _{=-{\frac {\pi ^{2}}{4}}\log 2+{\frac {13}{8}}\zeta (3)}\end{aligned}}} ∫01artanhx⋅logxx(1−x)(1+x)dx=−38ζ(3)−π216log2+14ζ(3)+132ζ(3)−12ζ(3)−14ζ(3)−π216log2+1332ζ(3)=−716ζ(3)−π28log2{\displaystyle \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}