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I'm trying to evaluate the following integral, which arised while attempting to find the sum of a series : $$\int_{0}^{1} \frac{\ln(x)}{x-1} \ln(1+\sqrt{x})\text{d}x$$

I've tried unsuccessfully some substitutions, integration by parts, feynman integration... I'm not familiar with more advanced integration techniques like residues theorem etc. ,so maybe that's the way to go. Any hint, solution or partial solution would be nice !

Harmonic Sun
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    Not relevant, but this integral is not "simple" at the first glimpse… – xbh Aug 18 '18 at 05:43
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    Mathematica gives the value $$\frac{1}{4}(\pi^2 \log 4 - 9 \zeta(3)),$$ which suggests that one might be able to rewrite the integrand in a clever way, but I don't have a solution at this time. – heropup Aug 18 '18 at 05:52
  • @heropup. What is funny is that the antiderivative seems to exist in terms of polylogarithms. Would you check that Mathematica gives it ? Thanks. – Claude Leibovici Aug 18 '18 at 06:07
  • When I saw the zeta(3) in the value you just gave me, I immediatly thought that there's ought to be some trilogarithm hidden in there... Thank you by the way. I don't have a mathematica lisence though, and wolfram is unable to proceed it – Harmonic Sun Aug 18 '18 at 06:08
  • @ClaudeLeibovici Yes, it does, but it is quite lengthy and its evaluation at the endpoints requires some delicacy as it appears to involve taking limits. I think an approach specific to the definite integral is more likely to be fruitful. – heropup Aug 18 '18 at 06:11

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You can reduce your integral to two integrals which have already been solved on this site:

Let $x=t^2$, perform a partial fraction decomposition and integrate by parts in the second integral to find \begin{align} I &\equiv \int\limits_0^1 \frac{-\ln(x) \ln(1+\sqrt{x})}{1-x} \, \mathrm{d}x \\ &= 4 \int \limits_0^1 [-\ln(t) \ln(1+t)] \frac{t}{(1-t)(1+t)} \, \mathrm{d}t \\ &= 2 \int \limits_0^1 \frac{-\ln(t) \ln(1+t)}{1-t} \, \mathrm{d} t - 2 \int \limits_0^1 \frac{-\ln(t) \ln(1+t)}{1+t} \, \mathrm{d} t \\ &= 2 \int \limits_0^1 \frac{-\ln(t) \ln(1+t)}{1-t} \, \mathrm{d} t - \int \limits_0^1 \frac{\ln^2 (1+t)}{t} \, \mathrm{d} t \\ &\equiv 2 I_1 - I_2 \, . \end{align}

In this question $I_2 = \frac{\zeta(3)}{4}$ is proved (FDP's answer does not use contour integration or the polylogarithm). $I_1 = \frac{\pi^2}{4} \ln(2) - \zeta(3)$ is derived here (the answer relies on contour integration though and it would be nice to have a simpler proof). Thus we obtain the result suggested in the comments: $$ I = 2 \left(\frac{\pi^2}{4} \ln(2) - \zeta(3)\right) - \frac{\zeta(3)}{4} = \frac{\pi^2}{2} \ln(2) - \frac{9}{4} \zeta(3) \, . $$

ComplexYetTrivial
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