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Question Number 157251 by mnjuly1970 last updated on 21/Oct/21

$$$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$...calculation...$$$$\mathrm{\Omega }:={\int }_0^1\frac{{tanh}^{-1}\left(\sqrt{x}\right)}{x}dx\stackrel{?}{=}\frac{{\pi }^2}{4}$$$$-------------$$$$\mathrm{\Omega }:\stackrel{\sqrt{x}=t}{=}2{\int }_0^1\frac{{tanh}^{-1}\left(t\right)}{t}dt$$$$:\stackrel{\{{tanh}^{-1}\left(t\right)=\frac{1}{2}ln\left(\frac{1+t}{1-t}\right)\}}{=}{\int }_0^1\frac{ln(1+t)-ln(1-t)}{t}dt$$$$:=-{\mathrm{Li}}_2(-1)+{\mathrm{Li}}_2\left(1\right)$$$$:\stackrel{\{{\mathrm{Li}}_2\left(z\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{z^n}{n^2}\}}{=}\eta \left(2\right)+\zeta \left(2\right)$$$$:=\frac{{\pi }^2}{12}+\frac{{\pi }^2}{6}=\frac{{\pi }^2}{4}◼m.n$$$$$$

Answered by qaz last updated on 21/Oct/21

$${\int }_0^1\frac{\tanh {}^{-1}\left(\sqrt{\mathrm{x}}\right)}{\mathrm{x}}\mathrm{dx}$$$$=2{\int }_0^1\frac{\tanh {}^{-1}\mathrm{x}}{\mathrm{x}}\mathrm{dx}$$$$=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2\mathrm{n}+1}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}$$$$=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2\mathrm{n}+1)}^2}$$$$=2(1-2^{-2})\zeta \left(2\right)$$$$=\frac{{\pi }^2}{4}$$

Commented by mnjuly1970 last updated on 21/Oct/21

$$thanksalotsirqaz$$

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