29

How can I prove in an elementary way that: $$\int _0^{\pi /2}x\operatorname{Li}_3\left(\sin \left(x\right)\right)\:\mathrm{d}x=\int _0^1\frac{\arcsin \left(x\right)\operatorname{Li}_3\left(x\right)}{\sqrt{1-x^2}}\:\mathrm{d}x$$ $$=\frac{5}{8}\zeta(2)\zeta(3)-\frac{217}{256}\zeta(5)-\frac{55}{64}\log \left(2\right)\zeta(4)+\frac{1}{3}\log ^3\left(2\right)\zeta(2)-\frac{1}{48}\log ^5\left(2\right)+\frac{5}{2}\operatorname{Li}_5\left(\frac{1}{2}\right),$$ where $\zeta$ is the Riemann zeta function and $\operatorname{Li}_n$ denotes the polylogarithm.

The closed form is simple enough; however, I am not able to reduce it to Euler sums which yield similar closed forms. I tried converting the trilogarithm into an integral as follows: $$\int _0^{\pi /2}x\operatorname{Li}_3\left(\sin \left(x\right)\right)\:\mathrm{d}x=\frac{1}{2}\int _0^1\log ^2\left(t\right)\int _0^{\pi /2}\frac{x\sin \left(x\right)}{1-\sin \left(x\right)t}\:\mathrm{d}x\:\mathrm{d}t.$$ I tried some substitutions but couldn't obtain anything useful from this. Furthermore, I converted the integral into series but the Euler sums which appear subsequently are challenging. For example: $$\int _0^1\frac{\arcsin \left(x\right)\operatorname{Li}_3\left(x\right)}{\sqrt{1-x^2}}\:\mathrm{d}x$$ $$=\int _0^1\frac{1}{2}\sum _{n=1}^{\infty }\frac{4^n}{n\binom{2n}{n}}x^{2n-1}\operatorname{Li}_3\left(x\right)\:\mathrm{d}x=\frac{1}{2}\sum _{n=1}^{\infty }\frac{4^n}{n\binom{2n}{n}}\int _0^1x^{2n-1}\operatorname{Li}_3\left(x\right)\:\mathrm{d}x$$ $$=\frac{1}{2}\sum _{n=1}^{\infty }\frac{4^n}{n\binom{2n}{n}}\left(\frac{1}{8}\frac{H_{2n}}{n^3}-\frac{1}{4}\zeta(2)\frac{1}{n^2}+\frac{1}{2}\zeta(3)\frac{1}{n}\right)$$ $$=\frac{1}{16}\sum _{n=1}^{\infty }\frac{4^nH_{2n}}{n^4\binom{2n}{n}}-\frac{1}{8}\zeta(2)\sum _{n=1}^{\infty }\frac{4^n}{n^3\binom{2n}{n}}+\frac{1}{4}\zeta(3)\sum _{n=1}^{\infty }\frac{4^n}{n^2\binom{2n}{n}}.$$ The leftmost series is very difficult whilst the rest are quite simple.

Saul
  • 254
  • I know the value of a lower weight sum, $\sum {n=1}^{\infty }\frac{4^nH{2n}}{n^3\binom{2n}{n}}$. Not sure if it's useful. – Ali Olaikhan Mar 12 '25 at 01:43
  • @AliShadhar I'm afraid it isn't. – Saul Mar 12 '25 at 04:44
  • It looks like we can write $x=\tan^{-1}(\tan x)$ and write $\tan^{-1}$ as its integral form and evaluate this double integral. Nice question too –  Mar 12 '25 at 07:11
  • @AliShadhar Can you modify the way you solved the sum you know to this one? It seems very close... – Kamal Saleh Mar 12 '25 at 19:19
  • @KamalSaleh His way involves integrating by parts some special integrals, reducing them to known/more manageable integrals. This isn't possible with the more advanced version since integration by parts would not achieve any reduction. – Saul Mar 12 '25 at 20:46
  • An extension: $$\int_0^{\pi/2} x \text{Li}_4(\sin x) dx$$ equals $$-\frac{224}{69} \Re\left(\text{Li}_6\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{1}{12} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{190 \text{Li}_6\left(\frac{1}{2}\right)}{69}-\frac{7 \zeta (3)^2}{128}-\frac{1}{32} \pi ^2 \zeta (3) \log (2)+\frac{6677 \pi ^6}{5806080}+\frac{373 \log ^6(2)}{99360}-\frac{863 \pi ^2 \log ^4(2)}{79488}+\frac{4471 \pi ^4 \log ^2(2)}{1589760}.$$ One could also try $$\int_0^{\pi/2} x (\text{Li}_2(\sin x))^2 dx$$ – pisco Mar 13 '25 at 07:26
  • Note that $$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=-\frac14\int_0^1\frac{\arcsin^2(x)\ln x}{x}dx$$ – Maxime Jaccon Mar 14 '25 at 02:13
  • I know a way to evaluate the left most sum, but the result will be in terms of hypergeometric functions, if that helps. One has to know some reduction formula, which is elusive to me at the moment. – Erosannin Mar 14 '25 at 05:16
  • @MaximeJaccon Unfortunately, that fact is irrelevant for the leftmost harmonic series. – Saul Mar 14 '25 at 06:41
  • @Saul May I know how you got the expression $\int _0^{\pi /2}x\operatorname{Li}_3\left(\sin \left(x\right)\right):\mathrm{d}x=\frac{1}{2}\int _0^1\log ^2\left(t\right)\int _0^{\pi /2}\frac{x\sin \left(x\right)}{1-\sin \left(x\right)t}:\mathrm{d}x:\mathrm{d}t$ ? I just got introduced to polylogs because of your post. – vishalnaakar25 Mar 24 '25 at 06:38
  • @vishalnaakar25 I used that $\displaystyle \int 0^1\frac{x\ln ^n\left(t\right)}{1-xt}:\mathrm{d}t=\left(-1\right)^nn!\operatorname{Li}{n+1}\left(x\right)$ – Saul Mar 27 '25 at 01:53
  • @Saul Thank you ! That's an interesting relation. I just found it from wiki in the polylog - Integral representation subsection as well. – vishalnaakar25 Mar 27 '25 at 04:39

1 Answers1

6

You already got $$\int _0^{\pi /2}x\operatorname{Li}_3\left(\sin \left(x\right)\right)\:\mathrm{d}x=\frac{1}{16}\underbrace{\sum _{n=1}^{\infty }\frac{4^nH_{2n}}{n^4\binom{2n}{n}}}_{A}-\frac{1}{8}\zeta(2)\underbrace{\sum _{n=1}^{\infty }\frac{4^n}{n^3\binom{2n}{n}}}_{B}+\frac{1}{4}\zeta(3)\underbrace{\sum _{n=1}^{\infty }\frac{4^n}{n^2\binom{2n}{n}}}_{C}.$$

Here, we have $B=\pi^2\log(2)-\frac{7}{2}\zeta(3)$ and $C=\frac{\pi^2}{2}$ (see for example here).

For $A$, I found this paper which studies some Euler-Apéry-type series involving central binomial coefficients and (generalized) harmonic numbers.

It follows from Theorem 3.6. (page 18) that $$A=40\operatorname{Li}_5\left(\frac{1}{2}\right)-\frac{217}{16}\zeta(5)-9\zeta(3)\zeta(2)+\frac{65}{4}\zeta(4)\log(2)+\frac{16}{3}\zeta(2)\log^3(2)-\frac 13\log^5(2)\tag1$$ (see also 4. Further remarks (page 20~21).)

Therefore, we finally get $$\int _0^{\pi /2}x\operatorname{Li}_3\left(\sin \left(x\right)\right)\:\mathrm{d}x=\int _0^1\frac{\arcsin \left(x\right)\operatorname{Li}_3\left(x\right)}{\sqrt{1-x^2}}\:\mathrm{d}x$$

$$=-\frac{1}{8}\zeta(2)\zeta(3)-\frac{217}{256}\zeta(5)+\frac{65}{64}\log \left(2\right)\zeta(4)+\frac{1}{3}\log ^3\left(2\right)\zeta(2)-\frac{1}{48}\log ^5\left(2\right)+\frac{5}{2}\operatorname{Li}_5\left(\frac{1}{2}\right)-\frac 18\log(2)\zeta(2)\pi^2+\frac 18\zeta(3)\pi^2$$

$$=-\frac{1}{8}\zeta(2)\zeta(3)-\frac{217}{256}\zeta(5)+\frac{65}{64}\log \left(2\right)\zeta(4)+\frac{1}{3}\log ^3\left(2\right)\zeta(2)-\frac{1}{48}\log ^5\left(2\right)+\frac{5}{2}\operatorname{Li}_5\left(\frac{1}{2}\right)-\frac 18\log(2)\times 15\zeta(4)+\frac 18\zeta(3)\times 6\zeta(2)$$

$$=\frac{5}{8}\zeta(2)\zeta(3)-\frac{217}{256}\zeta(5)-\frac{55}{64}\log \left(2\right)\zeta(4)+\frac{1}{3}\log ^3\left(2\right)\zeta(2)-\frac{1}{48}\log ^5\left(2\right)+\frac{5}{2}\operatorname{Li}_5\left(\frac{1}{2}\right)$$

mathlove
  • 151,597
  • 6
    I'm sorry but I would prefer a more elementary way about such integral. Preferably an approach based on real methods which does not utilize advanced MZV theory. – Saul Mar 15 '25 at 19:23
  • 10
    Perhaps you could clarify what type of answer you're looking for within the body of your question? As your question is currently written, this is a perfectly acceptable answer. Forcing others to chase a moving goalpost is just unreasonable. – teadawg1337 Mar 15 '25 at 20:10
  • 12
    I added "elementary" in the body now. Furthermore, given my attempts, it should be clear what kind of answer I am expecting. Thank you. – Saul Mar 15 '25 at 20:23
  • 4
    Woah! (+1) How did you find the paper? :) – Maxime Jaccon Mar 15 '25 at 20:31
  • 13
    Better call @Saul – Carlo Mar 16 '25 at 16:35