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

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{displaystyle int _{0}^{1}{frac {{ ext{artanh}}\,x\,\,log x}{x\,(1-x)\,(1+x)}}\,dx=-{frac {1}{16}}{Big (}7zeta (3)+2pi ^{2}log 2{Big )}}

Beweis

{displaystyle { ext{artanh}}\,x={frac {1}{2}}cdot log left({frac {1+x}{1-x}} ight)={frac {1}{2}}cdot {Big [}log(1+x)-log(1-x){Big ]}}

{displaystyle Rightarrow \,{ ext{artanh}}\,xcdot log x={frac {1}{2}}cdot log(1+x)log x-{frac {1}{2}}cdot log(1-x)log 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}}}


{displaystyle {egin{aligned}{frac {{ ext{artanh}}\,xcdot 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}}}



{displaystyle {egin{aligned}int _{0}^{1}{frac {{ ext{artanh}}\,xcdot 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}}}


{displaystyle int _{0}^{1}{frac {{ ext{artanh}}\,xcdot 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}