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Question Number 157278 by mnjuly1970 last updated on 21/Oct/21

Answered by mindispower last updated on 21/Oct/21

$${tanh}^{-}\left(x\right)=\sum _{m⩾0}\frac{x^{2m+1}}{2m+1}$$$${\int }_0^1\frac{ln(1-\sqrt{x}){tanh}^{-}\left(\sqrt{x}\right)}{x}dx$$$$={\int }_0^1\frac{ln(1-\sqrt{x})}{\sqrt{x}}{tanh}^{-}\left(\sqrt{x}\right).2d\sqrt{x}$$$$={\int }_0^1\frac{ln(1-y)}{y}\tan h^{-}\left(y\right)dy$$$$=2{\int }_0^1\sum _{m⩾0}\frac{1}{2m+1}ln(1-y)y^{2m}dy$$$$=\sum _{m⩾0}\frac{2}{2m+1}{\int }_0^{1}ln(1-y)y^{2m}dy$$$$=\sum _{m⩾0}\frac{2}{2m+1}{\partial }_a{\int }_0^1{(1-y)}^a{y}^{2m}dy{\mid }_{a=0}$$$$=\mathrm{\Sigma }\frac{2}{2m+1}{\partial }_a\beta (2m+1,1+a){\mid }_{a=0}$$$$=\sum _{m⩾0}\frac{2}{2m+1}\beta (2m+1,1)(\mathrm{\Psi }\left(1\right)-\mathrm{\Psi }(2+2m))$$$$=\sum _{m⩾1}\frac{2}{2m+1}.\frac{1}{2m+1}(-H_{2m+1})$$$$=\sum _{m⩾0}\frac{-2H_{2m+1}}{{(2m+1)}^2}$$$$$$$$$$$$$$

Commented by mnjuly1970 last updated on 22/Oct/21

$$thankyousomuchsirpower$$

Commented by mindispower last updated on 22/Oct/21

$$withePleasur$$

Answered by qaz last updated on 22/Oct/21

$${\int }_0^1\frac{\ln (1-\sqrt{\mathrm{x}}){\tanh }^{-1}\sqrt{\mathrm{x}}}{\mathrm{x}}\mathrm{dx}$$$$=2{\int }_0^1\frac{\ln (1-\mathrm{x}){\tanh }^{-1}\mathrm{x}}{\mathrm{x}}\mathrm{dx}$$$$=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2\mathrm{n}+1}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}}\ln (1-\mathrm{x})\mathrm{dx}$$$$=-2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_{2\mathrm{n}+1}}{{(2\mathrm{n}+1)}^2}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}({(-1)}^{\mathrm{n}}-1)\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2}$$$$------------$$$${\mathrm{H}}_{\mathrm{n}}={\int }_0^1\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}\mathrm{dx}=-(1-{\mathrm{x}}^{\mathrm{n}})\ln (1-\mathrm{x}){\mid }_0^1-\mathrm{n}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}=-\mathrm{n}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}$$$$\Rightarrow {\mathrm{H}}_{\mathrm{n}}=-\mathrm{n}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}$$

Commented by mnjuly1970 last updated on 22/Oct/21

$$thanksalotsirqaz$$

Answered by mnjuly1970 last updated on 22/Oct/21

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