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I need to evaluate $$\int_{1}^{\infty} \frac{\displaystyle{\operatorname{Re}\left ( \operatorname{Li}_3\left ( \frac{1+x}{2} \right ) \right ) \ln^2\left ( \frac{1+x}{2} \right ) }}{x(1+x^2)} \text{d}x.$$ $\operatorname{Li}_n(.)$ denotes the "Polylogarithms".
Numerical tests derived $$\int_{1}^{\infty} \frac{\displaystyle{\operatorname{Re}\left ( \operatorname{Li}_3\left ( \frac{1+x}{2} \right ) \right ) \ln^2\left ( \frac{1+x}{2} \right ) }}{x(1+x^2)} \text{d}x =\operatorname{Im}\left[\operatorname{Li}_3\left ( \frac{1+i}{2} \right ) \right]^2+\frac{1911}{4096}\zeta(3)^2 +\frac{413}{1536}\zeta(3)\ln^32 -\frac{721}{6144}\pi^2\zeta(3)\ln2-\frac{41}{1536}\pi^2\ln^42 +\frac{77}{12288}\pi^4\ln^22+\frac{7}{256}\ln^62.$$

I checked this $$\int_{1}^{\infty} \frac{\displaystyle{\operatorname{Re}\left ( \operatorname{Li}_4\left ( \frac{1+x}{2} \right ) \right ) \ln^3\left ( \frac{1+x}{2} \right ) }}{x(1+x^2)} \text{d}x =3\operatorname{Im} \left [ \operatorname{Li}_4\left ( \frac{1+i}{2} \right ) \right ]^2 -\frac{343\pi^4\operatorname{Li}_4\left ( \frac{1}{2} \right ) }{49152} +\frac{693\operatorname{Li}_4\left ( \frac{1}{2} \right )^2}{256} -\frac{5\operatorname{Li}_4\left ( \frac{1}{2} \right )}{256}\ln^42 +\frac{25\pi^2\operatorname{Li}_4\left ( \frac{1}{2} \right )}{2048}\ln^22 -\frac{117649\pi^8}{2831155200}-\frac{\ln^82}{3072} +\frac{5\pi^2\ln^62}{12288}-\frac{1061\pi^4\ln^42}{2949120} +\frac{343\pi^6\ln^22}{2359296}.$$

Setness Ramesory
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    Where did you find this monster? – Laxmi Narayan Bhandari Oct 16 '21 at 08:26
  • I suggest math overflow for this integral. – RAHUL Oct 16 '21 at 08:56
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    Yes, this is true. Such integral (i.e. level $4$ weight $6$) will be routinely evaluated by a new version of MZIntegrate, which will be released soon. But I think an more elementary way is possible judging on its simple result. – pisco Oct 16 '21 at 10:28
  • @pisco How to prove $\int_{2-\sqrt{3} }^{1} \frac{\log\left ( \frac{1-x}{\sqrt{2x} } \right ) \left [ \frac{\pi}{2}\log\left ( \frac{1+x}{\sqrt{2x} } \right ) -\operatorname{Ti}_2\left ( \frac{1+x}{\sqrt{2x} } \right ) \right ] }{1+x^2} \text{d}x=\frac{1}{2}C^2$? – Setness Ramesory Mar 02 '22 at 11:49
  • @SetnessRamesory No idea, this seems quite nontrivial. – pisco Mar 02 '22 at 14:19
  • @pisco Any way to show $\int_{0}^{\frac{\pi}{2} }\ln\Big[ K(\sin x)^2+K(\cos x)^2 \Big]\text{d}x =\pi\ln(\pi)$? – Setness Ramesory Mar 09 '22 at 10:56
  • @SetnessRamesory Your second integral can be done by contour integration. Where did you get these problems? – pisco Mar 15 '22 at 21:31
  • @pisco Which is the second integral? – Setness Ramesory Mar 19 '22 at 02:08
  • @SetnessRamesory $\int_{0}^{\frac{\pi}{2} }\ln\Big[ K(\sin x)^2+K(\cos x)^2 \Big]\text{d}x$ – pisco Mar 19 '22 at 09:15
  • @pisco Is it possible to compute $\int_{0}^{\infty}x^{s-1}\vartheta_2(q)^6\vartheta_4(q)^2 \text{d}x,q=e^{-\pi x}$? – Setness Ramesory Mar 25 '22 at 09:43
  • @SetnessRamesory Using technique similar to https://math.stackexchange.com/questions/3805720. If $\vartheta_2^6(2z) \vartheta_4^2(2z)$ is modular of weight $4$ to some congruence subgroup, and there is no cusp form, then your integral can be written as Dirichlet-L functions. If there are cusp form, then things become more difficult. – pisco Mar 25 '22 at 09:52
  • @pisco $\frac{3}{2} \int_{0}^{2} \frac{\arctan\left(\sqrt{2+x^2}\right) }{ (1+x^2)\sqrt{2+x^2} } \text{d}x +\int_{0}^{1} \frac{\arctan\left(\sqrt{5+x^2}\right)}{(4+x^2)\sqrt{5+x^2} } \text{d} x =\frac{\pi}{4}\left ( \arctan\left ( \frac{1}{2\sqrt{6} } \right ) +3\arctan\left ( \sqrt{\frac{2}{3} } \right ) \right ) -\frac{\pi^2}{16}$? – Setness Ramesory Apr 02 '22 at 10:47
  • @SetnessRamesory Yes, it's true. – pisco Apr 02 '22 at 12:04

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