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Question Number 97369 by  M±th+et+s last updated on 07/Jun/20

$${\int }_0^1\frac{{\left(ln\left(x\right)\right)}^2}{1+x^2}dx=\frac{{\pi }^3}{16}$$

Answered by maths mind last updated on 07/Jun/20

$$hellohappytocomback$$$${\int }_0^1\frac{{ln}^2\left(z\right)}{1+z^2}dz={\int }_1^{\mathrm{\infty }}\frac{{ln}^2\left(z\right)}{1+z^2}dz$$$$\Rightarrow {\int }_0^1\frac{{ln}^2\left(z\right)}{1+z^2}dz=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(z\right)}{1+z^2}dz$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{ln}^2\left(z\right)}{1+z^2}dz=2i\pi Res(\frac{{ln}^2\left(z\right)}{1+z^2},z=i)$$$$\Rightarrow {\int }_0^{+\mathrm{\infty }}\frac{{ln}^2\left(z\right)}{1+z^2}dz+{\int }_{-\mathrm{\infty }}^0\frac{{ln}^2\left(z\right)}{1+z^2}dz$$$$={\int }_0^{+\mathrm{\infty }}\frac{{ln}^2\left(z\right)+{ln}^2\left(z\right)-{\pi }^2+2i\pi ln\left(z\right)}{1+z^2}dz=2i\pi .\frac{{ln}^2\left(i\right)}{2i}=\pi .-\frac{{\pi }^2}{4}$$$$\iff 2{\int }_0^{+\mathrm{\infty }}\frac{{ln}^2\left(z\right)}{1+z^2}dz-{\pi }^2{\int }_0^{+\mathrm{\infty }}\frac{dz}{1+z^2}=-\frac{{\pi }^3}{4}$$$$\Rightarrow 2{\int }_0^{+\mathrm{\infty }}\frac{{ln}^2\left(z\right)}{1+z^2}dz-\frac{{\pi }^3}{2}=-\frac{{\pi }^3}{4}\iff {\int }_0^{+\mathrm{\infty }}\frac{{ln}^2\left(z\right)}{1+z^2}=2{\int }_0^1\frac{{ln}^2\left(z\right)}{1+z^2}dz=\frac{{\pi }^3}{8}$$$$\Rightarrow {\int }_0^1\frac{{ln}^2\left(z\right)}{1+z^2}dz=\frac{{\pi }^3}{16}$$

Commented by  M±th+et+s last updated on 07/Jun/20

$$howareyousirwellllllcomeeeeeback$$$$veryhappytoseeyouagain$$

Commented by maths mind last updated on 07/Jun/20

$$imfinthanxihopyoutoo$$$$$$$$$$

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