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This questions is related to my previous one.

I am interested in a explicit evaluation in terms of Euler sums for $$ \int_{0}^{\pi/4}\text{Li}_3(\cos^2\theta)\,d\theta = \frac{1}{2}\int_{0}^{1/2}\frac{\text{Li}_3(1-z)}{\sqrt{z(1-z)}}\,dz.$$

It is not difficult to show that $$ \int_{0}^{\color{red}{1}}\frac{\text{Li}_3(1-z)}{\sqrt{z(1-z)}}\,dz =-\frac{\pi^3}{3}\log(2)+\frac{4\pi}{3}\log^3(2)+2\pi\zeta(3)\tag{A}$$ but I have not managed to make a wise use of the trilogarithm functional identities for computing $$ \int_{0}^{1/2}\frac{\text{Li}_3(z)}{\sqrt{z(1-z)}}\,dz\stackrel{\text{IBP}}{\longrightarrow}\int_{0}^{\pi/4}\theta\cot(\theta)\text{Li}_2(\sin^2\theta)\,d\theta \quad\text{or}\quad\int_{0}^{1/2}\frac{\text{Li}_3(z)-\text{Li}_3(1-z)}{\sqrt{z(1-z)}}\,dz ,$$ which would have solved the problem. One might need substantial extensions of the result about $\mathcal{I}(a,b)$ proved here by nospoon. I am expecting the integral above to be related with Euler sums with (total) weight five. Maybe the shifted-Fourier-Chebyshev expansion of $\text{Li}_3(x)$ over $(0,1)$ is already known in the literature, but I have not been able to find it.

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  • $\begingroup$ There is no tag "polylogarithms"? (I cannot understand the classification in (hypergeometric-function)...) $\endgroup$
    – dan_fulea
    Commented May 29, 2018 at 0:15
  • $\begingroup$ @dan_fulea: You are right, I forgot to add it. The hypergeometric tag is due to the fact that such integral gives a closed form for a peculiar $\phantom{}_5 F_4$ with half-integer parameters. $\endgroup$
    – Jack D'Aurizio
    Commented May 29, 2018 at 0:16

2 Answers 2

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This is meant to be just a comment (too large to add it in the comments section)

$$I=\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{2}{3}\sum_{n=2}^{\infty}(-1)^{n-1}\frac{n(\overbrace{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)})}^{G1}-3 (\overbrace{H_n^2+ H_n^{(2)})}^{G2}}{2n(2n-1)(2n-2)}.$$

A wise next move would be to split the series into 2 series using the groups $G1$ and $G2$ and try to calculate the resulting series like that. In the mathematical literature there are nice, useful representations for those harmonic numbers groups.

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    $\begingroup$ The series related to G2 are manageable. Then we may perform a partial fraction decomposition and exploit the results of Xu, Yang and Zhang (Explicit evaluation of quadratic Euler sums) to compute two other pieces. The hardcore part is the evaluation of $$ \sum_{n\geq 0}(-1)^n \frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{2n-1} $$ which does not seem to be covered by classical results. $\endgroup$
    – Jack D'Aurizio
    Commented May 29, 2018 at 16:29
  • $\begingroup$ On the other hand such sum is related to $\int_{0}^{\pi/4}\log^3(\cos\theta)\,d\theta$ which has been computed by nospoon. Can you provide the derivation of the identity above in full detail? I believe I might be able to turn it into a complete solution. (+1) in the meanwhile. $\endgroup$
    – Jack D'Aurizio
    Commented May 29, 2018 at 16:31
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    $\begingroup$ @JackD'Aurizio I checked all the integrals behind the scene. I think all of them can be calculated by combining real and complex methods, but there is some work to do there. Before making any update I want to investigate the possibility of getting major shortcuts. $\endgroup$
    – user 1591719
    Commented May 29, 2018 at 17:25
  • $\begingroup$ Perfectly fine. You are probably already aware of it, but consider that the integrals $\int_{0}^{\pi/4}\log^\mu(\cos t)\,dt$ and $\int_{0}^{\pi/4}\log^\mu(\sin t)\,dt$ for $\mu\in\{0,1,2,3\}$ are known, even if WA does not return them in a nice closed form. $\endgroup$
    – Jack D'Aurizio
    Commented May 29, 2018 at 17:57
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By brutally evaluating such poly-logsine integrals, or those quadratic Euler sums @user 1591719 proposed, one have

  • $\int_0^{\frac{1}{2}} \frac{\text{Li}_3(z)}{\sqrt{z (1-z)}} \, dz=-\frac{\pi ^2 C}{6}+C \log ^2(2)-56 \Im(\text{Li}_4(1+i))+4 \log (2) \Im(\text{Li}_3(1+i))+\pi \zeta (3)+\pi \log ^3(2)+\frac{17}{24} \pi ^3 \log (2)+\frac{7 \psi ^{(3)}\left(\frac{1}{4}\right)}{256}-\frac{7 \psi ^{(3)}\left(\frac{3}{4}\right)}{256}$
  • $\int_0^{\frac{1}{2}} \frac{\text{Li}_3(1-z)}{\sqrt{z (1-z)}} \, dz=\frac{\pi ^2 C}{6}-C \log ^2(2)+56 \Im(\text{Li}_4(1+i))-4 \log (2) \Im(\text{Li}_3(1+i))+\pi \zeta (3)+\frac{1}{3} \pi \log ^3(2)-\frac{25}{24} \pi ^3 \log (2)-\frac{7 \psi ^{(3)}\left(\frac{1}{4}\right)}{256}+\frac{7 \psi ^{(3)}\left(\frac{3}{4}\right)}{256}$

See here for the methods of calculating quadratic Euler sums. This offers a far more systematic and advanced solution.

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    $\begingroup$ Much appreciated! (+1) and accepted. $\endgroup$
    – Jack D'Aurizio
    Commented Jun 20, 2020 at 9:49

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