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Question Number 147262 by mnjuly1970 last updated on 19/Jul/21

$$$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$$$$$\mathrm{I}:={\int }_0^1{\mathrm{Li}}_2\left(x^2\right)dx=?$$$$$$$$$$

Answered by qaz last updated on 19/Jul/21

$${\int }_0^1{\mathrm{Li}}_2\left({\mathrm{x}}^2\right)\mathrm{dx}$$$$={\mathrm{xLi}}_2\left({\mathrm{x}}^2\right){\mid }_0^1+2{\int }_0^1\ln (1-{\mathrm{x}}^2)\mathrm{dx}$$$$=\frac{{\pi }^2}{6}+2{\int }_0^1(\mathrm{lnx}+\ln (1+\mathrm{x}))\mathrm{dx}$$$$=\frac{{\pi }^2}{6}-4+4\ln 2$$$$-----------$$$${\int }_0^1{\mathrm{Li}}_2\left({\mathrm{x}}^2\right)\mathrm{dx}$$$$=-{\int }_0^1{\int }_0^{{\mathrm{x}}^2}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dtdx}$$$$=-\mathrm{x}{\int }_0^{{\mathrm{x}}^2}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}{\mid }_0^1+{\int }_0^1\mathrm{x}\cdot 2\mathrm{x}\frac{\ln (1-{\mathrm{x}}^2)}{{\mathrm{x}}^2}\mathrm{dx}$$$$=\frac{{\pi }^2}{6}+2{\int }_0^1\ln (1-{\mathrm{x}}^2)\mathrm{dt}$$$$=\frac{{\pi }^2}{6}+2{\int }_0^1(\ln (1-\mathrm{x})+\ln (1+\mathrm{x}))\mathrm{dx}$$$$=\frac{{\pi }^2}{6}-4+4\ln 2$$

Commented by mnjuly1970 last updated on 19/Jul/21

$$thanksalotsirqaz...verynice...$$

Answered by Olaf_Thorendsen last updated on 19/Jul/21

$$\mathrm{I}={\int }_0^1{\mathrm{Li}}_2\left(x^2\right)dx$$$$\mathrm{I}={\int }_0^1\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{\left(x^2\right)}^k}{k^2}\mathrm{d}x$$$$\mathrm{I}={\int }_0^1\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^{2k}}{k^2}\mathrm{d}x$$$$\mathrm{I}={\left[\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^{2k+1}}{k^2(2k+1)}\right]}_0^1$$$$\mathrm{I}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2(2k+1)}$$$$\mathrm{I}=\frac{1}{2}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2(k+\frac{1}{2})}$$$$\mathrm{I}=\underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{k^2}-\frac{1}{k(k+\frac{1}{2})})$$$$\mathrm{I}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}-\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(k+1)(k+\frac{3}{2})}$$$$\mathrm{I}=\frac{{\pi }^2}{6}-\frac{\psi \left(\frac{3}{2}\right)-\psi \left(1\right)}{\frac{3}{2}-1}$$$$\psi \left(1\right)=-\gamma \mathrm{and}\psi \left(\frac{1}{2}\right)=-2\ln 2-\gamma$$$$\psi (z+1)=\psi \left(z\right)+\frac{1}{z}$$$$\Rightarrow \psi \left(\frac{3}{2}\right)=\psi \left(\frac{1}{2}\right)+2=-2\ln 2-\gamma +2$$$$\mathrm{I}=\frac{{\pi }^2}{6}-\frac{-2\ln 2-\gamma +2+\gamma }{\frac{1}{2}}$$$$\mathrm{I}=\frac{{\pi }^2}{6}+4\ln 2-4$$

Commented by mnjuly1970 last updated on 19/Jul/21

$$gratefulsirolaf..excellent...$$

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