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

6天前 4次浏览

##### 0.1Bearbeiten
{displaystyle int _{0}^{1}{frac {{text{artanh}},x,,log x}{x,(1-x),(1+x)}},dx=-{frac {1}{16}}{Big (}7zeta (3)+2pi ^{2}log 2{Big )}}
Beweis

{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 ]}}

{displaystyle Rightarrow ,{text{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 {begin{aligned}{frac {{text{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 {begin{aligned}int _{0}^{1}{frac {{text{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 {{text{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}