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Question Number 141319 by mnjuly1970 last updated on 18/May/21

$$$$$$prove::$$$$\mathrm{\Omega }:={\int }_0^1\frac{{ln}^2\left(x\right)}{1-x^4}dx=\frac{{\pi }^3}{32}+\frac{7}{8}\zeta \left(3\right)..$$

Answered by qaz last updated on 17/May/21

$${\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(x\right)}{1-x^4}dx$$$$={\int }_0^1\frac{{ln}^2\left(x\right)}{1-x^4}dx+{\int }_1^{\mathrm{\infty }}\frac{{ln}^2\left(x\right)}{1-x^4}dx$$$$={\int }_0^1\frac{{ln}^{2}x}{1-x^4}dx+{\int }_0^1\frac{x^2{ln}^{2}x}{x^4-1}dx$$$$={\int }_0^1\frac{{ln}^{2}x}{1+x^2}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{y^2{e}^{-y}}{1+e^{-2y}}dy$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^{\mathrm{\infty }}{y}^2{e}^{-(2n+1)y}dy$$$$=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^3}$$$$=\frac{{\pi }^3}{16}$$

Answered by Dwaipayan Shikari last updated on 17/May/21

$${\int }_0^{\mathrm{\infty }}\frac{{log}^2\left(x\right)}{1-x^4}dx$$$$={\int }_1^{\mathrm{\infty }}\frac{{log}^2\left(x\right)}{1-x^4}dx+{\int }_0^1\frac{{log}^2\left(x\right)}{1-x^4}dx\frac{1}{x}=i$$$$={\int }_0^1\frac{{log}^2\left(u\right)u^2}{u^4-1}+{\int }_0^1\frac{{log}^2\left(x\right)}{1-x^4}dx{x}^4=m{u}^4=k$$$$=-\frac{1}{64}{\int }_0^1\frac{{log}^2\left(k\right)k^{-1/4}}{1-k}+\frac{1}{64}{\int }_0^1\frac{{log}^2\left(m\right)m^{-3/4}}{1-m}dm$$$$=\frac{1}{64}({\psi }^{″}\left(\frac{3}{4}\right)-{\psi }^{″}\left(\frac{1}{4}\right))=\frac{{\pi }^3}{16}$$$$\psi (1-a)-\psi \left(a\right)=\pi cot\left(\pi a\right)$$$${\psi }^{\prime }(1-a)+{\psi }^{\prime }\left(a\right)={\pi }^2{csc}^2\left(\pi a\right)$$$${\psi }^{″}(1-a)-{\psi }^{″}\left(a\right)=2{\pi }^3{csc}^2\left(\pi a\right)cot\left(\pi a\right)$$$${\psi }^{″}\left(\frac{3}{4}\right)-{\psi }^{″}\left(\frac{1}{4}\right)=4{\pi }^3$$

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