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Question Number 157196 by MathSh last updated on 20/Oct/21

Answered by mindispower last updated on 21/Oct/21

$$\frac{{\pi }^3}{64}ln\left(2\right)-3{\int }_0^1\frac{ln(1+x)}{1+x^2}{\tan }^{-1}{\left(x\right)}^{2}dx$$$$=-3{\int }_0^{\frac{\pi }{4}}ln(1+tg\left(t\right))t^{2}dtt$$$${\int }_0^{\frac{\pi }{4}}ln(1+tg\left(t\right))t^{2}dt={\int }_0^{\frac{\pi }{4}}ln(1+tg(\frac{\pi }{4}-t)){(\frac{\pi }{4}-t)}^2$$$$\Rightarrow 2{\int }_0^{\frac{\pi }{4}}ln(1+tg\left(t\right))t^{2}dt={\int }_0^{\frac{\pi }{4}}ln\left(2\right)t^2-\frac{{\pi }^2}{16}{\int }_0^{\frac{\pi }{4}}ln(1+tan\left(t\right))dt$$$$+\frac{\pi }{2}{\int }_0^{\frac{\pi }{4}}tln(1+tg\left(t\right))dt$$$$=A-\frac{{\pi }^2}{16}B+\frac{\pi }{2}C$$$$A=\frac{ln\left(2\right)}{3}\left(\frac{{\pi }^3}{64}\right)$$$$C={\int }_0^{\frac{\pi }{4}}ln\left(\frac{\sqrt{2}sin(\frac{\pi }{4}+x)}{cos\left(x\right)}\right)tdt$$$$=-2{\int }_0^{\frac{\pi }{4}}ln\left(cos\left(t\right)\right)td\underset{=Y}{t}+\frac{ln\left(\sqrt{2}\right)}{2}.\frac{\pi }{4}$$$$Y=2{\int }_0^{\frac{\pi }{4}}\sum _{k⩾1}{(-1)}^k\frac{cos\left(2kx\right)}{k}xdx+2{\int }_0^{\frac{\pi }{4}}ln\left(2\right)tdt$$$$=2\sum _{k⩾1}\frac{{(-1)}^k}{k}[\frac{\pi }{8k}sin\left(\frac{k\pi }{2}\right)+\frac{cos\left(\frac{k\pi }{2}\right)-1}{4k^2}]+\frac{ln\left(2\right){\pi }^2}{16}$$$$=-2\sum _{k⩾0}\frac{{(-1)}^k\pi }{8{(2k+1)}^2}-\frac{1}{2}\sum _{k⩾1}\frac{{(-1)}^k}{k^3}+\frac{1}{2}\sum _{k⩾1}\frac{{(-1)}^k}{8k^3}+\frac{ln\left(2\right){\pi }^2}{16}$$$$-2(\frac{\pi }{8}C-\frac{21}{128}\zeta \left(3\right)-\frac{{\pi }^2}{32}ln\left(2\right))$$$$=-2\left(\frac{16\pi C-21\zeta \left(3\right)-4{\pi }^{2}ln\left(2\right)}{128}\right)+\frac{\pi ln\left(2\right)}{32}$$$${\int }_0^{\frac{\pi }{4}}ln(1+tg\left(t\right))dt=B$$$$={\int }_0^{\frac{\pi }{4}}ln\left(\frac{2}{1+tg\left(t\right)}\right)=B$$$$\iff B=\frac{\pi }{8}ln\left(2\right)$$$$A=\frac{ln\left(2\right)}{3}.\frac{{\pi }^3}{64}$$$${\int }_0^{\frac{\pi }{4}}{t}^{2}ln(1+tan\left(t\right))dt=\frac{1}{2}(\frac{ln\left(2\right)}{3}.\frac{{\pi }^3}{64}-\frac{{\pi }^2}{16}\left(\frac{\pi ln\left(2\right)}{8}\right)-2\pi \left(\frac{16\pi C-21\zeta \left(3\right)-4{\pi }^{2}ln\left(2\right)}{128}\right)+\frac{{\pi }^2}{2}.\frac{ln\left(2\right)}{32})=\mathrm{\Omega }$$$${\int }_0^{1(}\frac{{{tan}^{-}\left(x\right))}^3}{1+x}dx=\frac{{\pi }^{3}ln\left(2\right)}{64}-3\mathrm{\Omega }$$$$$$$$$$

Commented by MathSh last updated on 21/Oct/21

$$\mathrm{Perfect}\mathrm{dear}\mathrm{Ser},\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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