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Question Number 163842 by mnjuly1970 last updated on 11/Jan/22

$$$$$$calculate$$$$$$$$\mathrm{\Omega }={\int }_0^1{\left(\frac{\mathcal{A}rctanh\left(x\right)}{x^{}}\right)}^{2}dx=?$$$$--m.n--$$

Answered by amin96 last updated on 11/Jan/22

$$\mathbf{\Omega }=\frac{1}{4}{\int }_0^1\frac{1}{{\mathbf{x}}^2}{\mathbf{ln}}^2\left(\frac{1+\mathbf{x}}{1-\mathbf{x}}\right)\mathbf{dx}\stackrel{\frac{1-\mathbf{x}}{1+\mathbf{x}}=\mathbf{t}}{=}\frac{1}{2}{\int }_0^1\frac{{\mathbf{ln}}^2\left(\mathbf{t}\right)}{{(1-\mathbf{t})}^2}\stackrel{\mathbf{IBP}}{=}$$$$\stackrel{\mathbf{IBP}}{=}\frac{1}{2}{\left[\frac{\mathbf{t}}{1-\mathbf{t}}{\mathbf{ln}}^2\mathbf{t}\right]}_0^1-\frac{1}{2}{\int }_0^1\frac{\mathbf{t}}{1-\mathbf{t}}\times \frac{2\mathbf{lnt}}{\mathbf{t}}\mathbf{dt}=-{\int }_0^1\frac{\mathbf{lnt}}{1-\mathbf{t}}\mathbf{dt}\stackrel{\mathbf{t}=1-\mathbf{t}}{=}$$$$\stackrel{\mathbf{t}=1-\mathbf{t}}{=}-{\int }_0^1\frac{\mathbf{ln}(1-\mathbf{t})}{\mathbf{t}}\mathbf{dx}=\underset{\mathbf{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathbf{n}}{\int }_0^1{\mathbf{t}}^{\mathbf{n}-1}\mathbf{dt}=\underset{\mathbf{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathbf{n}}^2}=\mathit{\zeta }\left(2\right)$$

Commented by mnjuly1970 last updated on 11/Jan/22

$$thanksalotsiramin...$$$$haveanicetime$$

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