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Question Number 164653 by mnjuly1970 last updated on 20/Jan/22

$$$$$$solve$$$$\mathit{\varphi }={\int }_0^1\frac{{\ln }^2\left(x\right).{tanh}^{-1}\left(x\right)}{x}dx=?$$$$\mathrm{\Omega }={\int }_0^1\frac{{\left({tanh}^{-1}\left(x\right)\right)}^2}{1+x}=?$$$$----$$

Answered by Lordose last updated on 20/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^1\frac{{\left({\tanh }^{-1}\left(\mathrm{x}\right)\right)}^2}{1+\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Omega }=\frac{1}{4}{\int }_0^1\frac{{\ln }^2\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)}{1+\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Omega }\stackrel{\mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}}}{=}\frac{1}{4}{\int }_0^1\frac{{\ln }^2\left(\mathrm{x}\right)}{1+\mathrm{x}}=\frac{1}{4}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}-1}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}{\ln }^2\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{\Omega }\stackrel{\mathbf{IBP}\times 2}{=}\frac{1}{2}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{{\mathrm{n}}^3}=\mathit{\eta }\left(3\right)=\frac{3}{8}\mathit{\zeta }\left(3\right)$$$$\mathbf{\Omega }=\frac{3}{8}\mathit{\zeta }\left(3\right)$$

Answered by Ar Brandon last updated on 20/Jan/22

$$$$$$\varphi ={\int }_0^1\frac{{\ln }^{2}x{\tanh }^{-1}\left(x\right)}{x}dx$$$$={\left[\frac{{\ln }^{3}x}{3}{\tanh }^{-1}\left(x\right)\right]}_0^1-\frac{1}{3}{\int }_0^1\frac{{\ln }^{3}x}{1-x^2}dx$$$$=-\frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{8}{\int }_0^1\frac{u^{-\frac{1}{2}}{\ln }^{3}u}{1-u}du=\frac{1}{48}{\psi }^{\left(3\right)}\left(\frac{1}{2}\right)$$$$=2(\zeta \left(4\right)-\frac{1}{16}\zeta \left(4\right))=\frac{15}{8}\zeta \left(4\right)=\frac{15}{720}{\pi }^4$$

Commented by mnjuly1970 last updated on 20/Jan/22

$$verhnice...thankyou$$$$sirbrandon...$$

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