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Question Number 166690 by mnjuly1970 last updated on 25/Feb/22

$$$$$$Calculate$$$$If,\mathit{\varphi }={\int }_0^1\frac{{tanh}^{-1}\left(x^3\right)}{x}dx=\alpha .\zeta \left(2\right)$$$$then,\alpha =?◼\mathcal{M}.\mathcal{N}$$$$-------$$$$$$

Answered by amin96 last updated on 25/Feb/22

$$\underset{―}{\mathbf{solution}:}{\mathbf{x}}^3=\mathbf{t}\frac{\mathbf{dt}}{\mathbf{dx}}=3{\mathbf{x}}^2=3{\mathbf{t}}^{\frac{2}{3}}\mathbf{dx}=\frac{\mathbf{dt}}{3{\mathbf{t}}^{\frac{2}{3}}}$$$$\mathit{\varphi }=\frac{1}{3}{\int }_0^1\frac{\mathbf{arctgh}\left(\mathbf{t}\right)}{\mathbf{t}}\mathbf{dt}=\frac{1}{6}{\int }_0^1\frac{\mathbf{ln}\left(\frac{1+\mathbf{t}}{1-\mathbf{t}}\right)}{\mathbf{t}}\mathbf{dt}=$$$$=\frac{1}{6}{\int }_0^1\frac{1}{\mathbf{t}}\mathbf{ln}\left(\frac{1+\mathbf{t}}{1-\mathbf{t}}\right)\mathbf{dt}=\frac{1}{3}{\int }_0^1\frac{1}{\mathbf{t}}\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathbf{t}}^{2\mathbf{n}+1}}{2\mathbf{n}+1}\mathbf{dt}=$$$$=\frac{1}{3}\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2\mathbf{n}+1}{\int }_0^1{\mathbf{t}}^{2\mathbf{n}}\mathbf{dt}=\frac{1}{3}\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2\mathbf{n}+1)}^2}=$$$$=\frac{1}{3}(1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots )=\frac{1}{3}(\mathit{\zeta }\left(2\right)-\frac{1}{4}\mathit{\zeta }\left(2\right))=$$$$=\frac{1}{4}\mathit{\zeta }\left(2\right)\mathit{\alpha }=\frac{1}{4}\mathit{b}\mathit{y}\mathit{M}.\mathit{A}$$

Commented by mnjuly1970 last updated on 25/Feb/22

$$gratefulsirAmin...excellent$$

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