Here there is an integral that I've found playing with Wolfram Alpha online calculator (thus to me is a curiosity that it has indefinite integral) $$\int\frac{\operatorname{Li}_2(x)}{1+\sqrt{x}}\,dx,\tag{1}$$ where the function in the numerator of the integrand is the polylogarithm

$$\operatorname{Li}_2(z)=\sum_{k=1}^\infty\frac{z^k}{k^2},\tag{2}$$ see the related MathWorld's article if you want to know more about this special function.

Question. To me seem difficult to get the indefinite integral that provide us Wolfram Alpha online calculator. But can you provide me the ideas or first calculations to calculate such indefinite integral? That is, imagine that I need to explain to a friend/colleague a draft about the strategy to get such indefinite integral. Then, what is the recipe that I need to explain him/her to justify from the top (without all tedious details) the indefinite integral? Many thanks.

When I was playing, before knowing such a closed form of the indefinite integral, my intention was to justify $$\int_0^1\frac{\operatorname{Li}_2(x)}{1+\sqrt{x}}\,dx.$$

I say these words to provide what are my intentions, I believe that this definite integral isn't special but I was interested in calculate it when I was asking to the mentioned CAS.

  • $\begingroup$ Many thanks @AdrianKeister $\endgroup$ – user243301 Apr 25 '18 at 19:53
  • $\begingroup$ Closed form of the definite integral exist by Zeta and derivative hypergeometric function. $\endgroup$ – Mariusz Iwaniuk Apr 25 '18 at 20:13
  • $\begingroup$ @MariuszIwaniuk: I would say that a closed form exists in terms of $\pi^2, \log(2)$ and $\zeta(3)$ only. $\endgroup$ – Jack D'Aurizio Apr 25 '18 at 20:19
  • $\begingroup$ Many thanks for your attention and help @MariuszIwaniuk $\endgroup$ – user243301 Apr 25 '18 at 20:35
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    $\begingroup$ @user243301 Will it suffice to find anti-derivative over $0\le x\le 1$? (The integrand is guaranteed to be real-valued for such $x$) $\endgroup$ – David H Apr 29 '18 at 11:01

A natural temptation is to remove the square root from the denominator of the integrand function by enforcing the substitution $x=u^2$, then expanding $\text{Li}_2(u^2)$ as a Maclaurin series and convert the whole thing into a combination of Euler sums, hopefully with a low weight. Indeed

$$ \int_{0}^{1}\frac{\text{Li}_2(x)}{1+\sqrt{x}}=2\int_{0}^{1}\frac{u}{1+u}\text{Li}_2(u^2)\,du =4\int_{0}^{1}\left[1-\frac{1}{1+u}\right]\cdot\left[\text{Li}_2(u)+\text{Li}_2(-u)\right]\,du$$ where $$ \int_{0}^{1}\text{Li}_2(u^2)\,du = -4+\frac{\pi^2}{6}+4\log(2)$$ is straightforward and $$ \int_{0}^{1}\frac{\text{Li}_2(u)}{1+u}\,du =\frac{\pi^2}{6}\log(2)-\frac{5}{8}\zeta(3),\qquad \int_{0}^{1}\frac{\text{Li}_2(-u)}{1+u}\,du =-\frac{\pi^2}{12}\log(2)+\frac{1}{4}\zeta(3)$$ have already been proved on MSE. The involved techniques are just integration by parts and the functional relations for the dilogarithm function (a function with the sense of humour, according to D.Zagier).

  • $\begingroup$ Many thanks mine was a just curiosity, these were my thoughts. And now I see that can be stated a so simple explanation, even if the corresponding indefinite integral seems complicated. Thus yes, this kind of integrals have sense of humour. $\endgroup$ – user243301 Apr 25 '18 at 20:32
  • $\begingroup$ Thus it's possible to get the definite integral by your remarks (change of variables, $\operatorname{Li}_2(u^2)=\operatorname{Li}_2(u)+\operatorname{Li}_2(-u)$), and remarkable definite integrals), but I am going to wait about if there are users providing me more remarks about the corresponding indefinite integral. Many thanks!!! $\endgroup$ – user243301 Apr 25 '18 at 20:44

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