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Question Number 164585 by mnjuly1970 last updated on 19/Jan/22

$$$$$$\mathcal{I}={\int }_0^1\frac{{\mathrm{Li}}_2\left(x\right)}{1+x}dx=?$$$$------$$

Answered by mindispower last updated on 19/Jan/22

$${\int }_0^1\frac{{Li}_2\left(x\right)}{1+x}dx=ln\left(2\right)\frac{{\pi }^2}{6}+{\int }_0^1\frac{ln(1-x)ln(1+x)}{x}{dx}_{=A}$$$$A=\frac{1}{4}{\int }_0^1\frac{{ln}^2(1-x^2)}{x}-\frac{{ln}^2\left(\frac{1-x}{1+x}\right)}{x}dx$$$$=\frac{1}{8}{\int }_0^1\frac{{ln}^2(1-y)}{y}dy-\frac{1}{2}{\int }_0^1\frac{{ln}^2\left(y\right)}{1-y^2}dy$$$$=\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{t^2{e}^{-t}}{1-e^{-t}}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^2{e}^{-t}}{1-e^{-2t}}dt$$$$=\frac{1}{8}\mathrm{\Gamma }\left(3\right)\zeta \left(3\right)-\frac{1}{2}\sum _{n⩾0}{\int }_0^{\mathrm{\infty }}{t}^2{e}^{-(1+2\mathit{n})\mathit{t}}\mathit{d}\mathit{t}$$$$=\frac{\mathit{\zeta }\left(3\right)}{4}-\sum _{n⩾0}\frac{\mathrm{\Gamma }\left(3\right)}{2{(2n+1)}^3}$$$$=\frac{\zeta \left(3\right)}{4}-\frac{7}{8}\zeta \left(3\right)=-\frac{5}{8}\zeta \left(3\right)$$$${\int }_0^1\frac{{Li}_2\left(x\right)}{1+x}dx=\frac{ln\left(2\right)}{6}{\pi }^2-\frac{5}{8}\zeta \left(3\right)$$$$$$

Commented by mnjuly1970 last updated on 19/Jan/22

$$thankhousomuchsirpower$$$$.verynicesolution..$$$$$$

Commented by mindispower last updated on 19/Jan/22

$$WithePleasurHaveaniceDaysir$$

Answered by Lordose last updated on 19/Jan/22

$$$$$$\mathrm{I}={\int }_0^1\frac{{\mathrm{Li}}_2\left(\mathrm{x}\right)}{1+\mathrm{x}}\mathrm{dx}$$$$\mathrm{I}\stackrel{\mathbf{IBP}}{=}{\mathrm{Li}}_2\left(\mathrm{x}\right)\ln (1+\mathrm{x}){\mid }_0^1+{\int }_0^1\frac{\ln (1-\mathrm{x})\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{I}=\ln \left(2\right)\mathit{\zeta }\left(2\right)+{\int }_0^1\frac{\ln (1-\mathrm{x})\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{J}=-\frac{1}{2}({\int }_0^1{\ln }^2\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\frac{\mathrm{dx}}{\mathrm{x}}-{\int }_0^1\frac{{\ln }^2(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}-{\int }_0^1\frac{{\ln }^2(1+\mathrm{x})}{\mathrm{x}}\mathrm{dx})$$$$\mathrm{J}=-\frac{1}{2}(\mathrm{A}-\mathrm{B}-\mathrm{C})$$$$\mathrm{A}\stackrel{\mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}}}{=}2{\int }_0^1\frac{{\ln }^2\left(\mathrm{x}\right)}{1-{\mathrm{x}}^2}\mathrm{dx}=2\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{2\mathrm{k}}{\ln }^2\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{A}\stackrel{\mathbf{IBP}\times 2}{=}4\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2\mathrm{k}+1)}^3}=4\left(\frac{7\mathit{\zeta }\left(3\right)}{8}\right)=\frac{7\mathit{\zeta }\left(3\right)}{2}$$$$\mathbf{A}=\frac{7\mathit{\zeta }\left(3\right)}{2}$$$$\mathrm{B}\stackrel{\mathrm{x}=1-\mathrm{x}}{=}{\int }_0^1\frac{{\ln }^2\left(\mathrm{x}\right)}{1-\mathrm{x}}\mathrm{dx}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{\mathrm{k}-1}{\ln }^2\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{B}\stackrel{\mathbf{IBP}\times 2}{=}2\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{k}}^3}=2\mathit{\zeta }\left(3\right)$$$$\mathbf{B}=2\mathit{\zeta }\left(3\right)$$$$\mathrm{C}=\frac{1}{4}\mathit{\zeta }\left(3\right)$$$$\mathrm{J}=-\frac{1}{2}(\frac{7\mathit{\zeta }\left(3\right)}{2}-2\mathit{\zeta }\left(3\right)-\frac{\mathit{\zeta }\left(3\right)}{4})=-\frac{5\mathit{\zeta }\left(3\right)}{8}$$$$\mathrm{I}=\ln \left(2\right)\mathit{\zeta }\left(2\right)+\frac{5\mathit{\zeta }\left(3\right)}{8}$$$$\mathbf{I}=\mathit{\zeta }\left(2\right)\mathbf{ln}\left(2\right)+\frac{5\mathit{\zeta }\left(3\right)}{8}$$

Commented by mnjuly1970 last updated on 19/Jan/22

$$thxalotsirlordos$$

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