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Question Number 162002 by mnjuly1970 last updated on 25/Dec/21

$$$$$$calculate$$$$$$$$\mathrm{\Omega }={\int }_0^1{\mathrm{Li}}_2(1-x^4)dx=?$$$$-----$$

Answered by Lordose last updated on 27/Dec/21

$$$$$$\mathrm{\Omega }={\int }_0^1{\mathbf{Li}}_2(1-{\mathrm{x}}^4)\mathrm{dx}$$$$\mathrm{\Omega }\stackrel{\mathrm{x}={\mathrm{x}}^{\frac{1}{4}}}{=}\frac{1}{4}{\int }_0^1{\mathrm{x}}^{\frac{1}{4}-1}{\mathbf{Li}}_2(1-\mathrm{x})\mathrm{dx}$$$$\mathrm{\Omega }\stackrel{\mathbf{IBP}}{=}\frac{1}{4}({4\mathrm{x}}^{\frac{1}{4}}{\mathbf{Li}}_2(1-\mathrm{x}){\mid }_0^1-4{\int }_0^1\frac{{\mathrm{x}}^{\frac{1}{4}}\log \left(\mathrm{x}\right)}{1-\mathrm{x}}\mathrm{dx})$$$$\mathrm{\Omega }=-\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}+\frac{5}{4}-1}\log \left(\mathrm{x}\right)\stackrel{\mathbf{IBP}\times 2}{=}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{n}+\frac{5}{4})}^2}$$$$\mathbf{N}.\mathbf{B}::{\mathit{\psi }}^{\left(\mathbf{m}\right)}\left(\mathbf{z}\right)={(-1)}^{\mathbf{m}+1}\mathbf{m}!\underset{\mathbf{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathbf{z}+\mathbf{k})}^{\mathbf{m}+1}}$$$$\mathbf{\Omega }={\mathit{\psi }}^{\left(1\right)}\left(\frac{5}{4}\right)=8\mathbf{G}+{\mathit{\pi }}^2-16$$$$\mathit{\varnothing }\mathbf{sE}$$

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