Zurück zu Bestimmte Integrale
artanh x = 1 2 ⋅ log ( 1 + x 1 − x ) = 1 2 ⋅ [ 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 ]}} ⇒ artanh x ⋅ log x = 1 2 ⋅ log ( 1 + x ) log x − 1 2 ⋅ log ( 1 − x ) log x {\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} 1 x ( 1 − x ) ( 1 + x ) = 1 x + 1 2 ⋅ 1 1 − x − 1 2 ⋅ 1 1 + 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}}} artanh x ⋅ log x x ( 1 − x ) ( 1 + x ) = + 1 2 ⋅ log ( 1 + x ) log x x + 1 4 ⋅ log ( 1 + x ) log x 1 − x − 1 4 ⋅ log ( 1 + x ) log x 1 + x − 1 2 ⋅ log ( 1 − x ) log x x − 1 4 ⋅ log ( 1 − x ) log x 1 − x + 1 4 ⋅ log ( 1 − x ) log x 1 + 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}}} ∫ 0 1 artanh x ⋅ log x x ( 1 − x ) ( 1 + x ) d x = + 1 2 ⋅ ∫ 0 1 log ( 1 + x ) log x x d x ⏟ = − 3 4 ζ ( 3 ) + 1 4 ⋅ ∫ 0 1 log ( 1 + x ) log x 1 − x d x ⏟ = − π 2 4 log 2 + ζ ( 3 ) − 1 4 ⋅ ∫ 0 1 log ( 1 + x ) log x 1 + x d x ⏟ − 1 8 ζ ( 3 ) − 1 2 ⋅ ∫ 0 1 log ( 1 − x ) log x x d x ⏟ = ζ ( 3 ) − 1 4 ⋅ ∫ 0 1 log ( 1 − x ) log x 1 − x d x ⏟ = ζ ( 3 ) + 1 4 ⋅ ∫ 0 1 log ( 1 − x ) log x 1 + x d x ⏟ = − π 2 4 log 2 + 13 8 ζ ( 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}}} ∫ 0 1 artanh x ⋅ log x x ( 1 − x ) ( 1 + x ) d x = − 3 8 ζ ( 3 ) − π 2 16 log 2 + 1 4 ζ ( 3 ) + 1 32 ζ ( 3 ) − 1 2 ζ ( 3 ) − 1 4 ζ ( 3 ) − π 2 16 log 2 + 13 32 ζ ( 3 ) = − 7 16 ζ ( 3 ) − π 2 8 log 2 {\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}
![{\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 ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ef4191d840f411d743b1ef69905bcbbc84e33a3)
