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I am not sure if there exists a closed form for

$$I=\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$$

which seems non-trivial.

I used the reflection and landen's identity, didn't help much.

In case you are curious how I came up with this integral:

From here we have

$$\arcsin^3(x)=6\sum_{k=1}^\infty\left[\sum_{m=0}^{k-1}\frac{1}{(2m-1)^2}\right]\frac{{2k\choose k}}{4^k}\frac{x^{2k+1}}{2k+1},\quad |x|<1$$

differentiate both sides with respect to $x$ we get

$$\frac{\arcsin^2(x)}{\sqrt{1-x^2}}=2\sum_{k=1}^\infty\left[\sum_{m=0}^{k-1}\frac{1}{(2m-1)^2}\right]\frac{{2k\choose k}}{4^k}x^{2k}$$

use $\sum_{m=0}^{k-1}\frac{1}{(2m-1)^2}=H_{2k}^{(2)}-\frac14H_k^{(2)}$ and replace $x$ by $\sqrt{x}$ we get the form

$$\frac{\arcsin^2(\sqrt{x})}{\sqrt{1-x}}=2\sum_{k=1}^\infty\left[H_{2k}^{(2)}-\frac14H_k^{(2)}\right]\frac{{2k\choose k}}{4^k}x^{k}$$

Divide both sides by $x$ then $\int_0^y$ we have

$$\int_0^y\frac{\arcsin^2(\sqrt{x})}{x\sqrt{1-x}}dx=2\sum_{k=1}^\infty\left[H_{2k}^{(2)}-\frac14H_k^{(2)}\right]\frac{{2k\choose k}}{4^k}\frac{y^{k}}{k}$$

Next, multiply both sides by $-\frac{\ln(1-y)}{y}$ then $\int_0^1$ and use $-\int_0^1 y^{k-1}\ln(1-y)dy=\frac{H_k}{k}$

$$2\sum_{k=1}^\infty\left[H_{2k}^{(2)}-\frac14H_k^{(2)}\right]\frac{{2k\choose k}}{4^k}\frac{H_k}{k^2}=-\int_0^1\int_0^y\frac{\arcsin^2(\sqrt{x})\ln(1-y)}{xy\sqrt{1-x}}dxdy$$

$$=\int_0^1\frac{\arcsin^2(\sqrt{x})}{x\sqrt{1-x}}\left(-\int_x^1\frac{\ln(1-y)}{y}dy\right)dx$$

$$=\int_0^1\frac{\arcsin^2(\sqrt{x})}{x\sqrt{1-x}}\left(\zeta(2)-\text{Li}_2(x)\right)dx$$

$$\overset{\sqrt{x}=\sin\theta}{=}2\int_0^{\pi/2}\frac{x^2}{\sin x}(\zeta(2)-\text{Li}_2(\sin^2x))dx$$

So we have

$$\sum_{k=1}^\infty\left[H_{2k}^{(2)}-\frac14H_k^{(2)}\right]\frac{{2k\choose k}}{4^k}\frac{H_k}{k^2}=\zeta(2)\int_0^{\pi/2}\frac{x^2}{\sin x}dx-\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$$

The first integral can be calculated by applying integration by parts then using the fourier series of $\ln(\tan\frac x2)$.

Another question is, clearly the two sums on the LHS are convergent as the denominator blows to infinity much faster than the numerator. But does there exist a closed form for each?

All methods are welcome. Thank you

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  • $\begingroup$ @picso what do you think? $\endgroup$
    – Ali Shadhar
    Commented Aug 28, 2020 at 19:46
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    $\begingroup$ Your integral is $$-64S+\frac{\pi ^3 C}{3}+64 \pi \Im(\text{Li}_4(\frac{1}{2}+\frac{i}{2}))+16 \pi \log (2) \Im(\text{Li}_3(\frac{1}{2}+\frac{i}{2}))-72 \pi \beta(4)+20 \text{Li}_5(\frac{1}{2})+20 \text{Li}_4(\frac{1}{2}) \log (2)-\frac{55 \pi ^2 \zeta (3)}{48}+\frac{373 \zeta (5)}{8}+\frac{2 \log ^5(2)}{3}-\frac{13}{18} \pi ^2 \log ^3(2)+\frac{9}{16} \pi ^4 \log (2)$$ where $S$ is the real part of $$\sum_{a>b>c\geq 1} \frac{i^c}{a^3 b c}$$ which possibly cannot be simplified, but there are hypergeometric representations. $\endgroup$
    – pisco
    Commented Aug 29, 2020 at 3:00
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    $\begingroup$ Getting rid of the $_pF_q$ is more or less equivalent to getting rid that triple sum. This could be possible, but any substantial simplification requires deep algebraic-geometric theory. For example, the sum $A=\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^5}$ familiar to you was widely believed not expressible in terms of $\text{Li}_n$, but in 2019, it was proved that ... $\endgroup$
    – pisco
    Commented Aug 29, 2020 at 4:03
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    $\begingroup$ $$A = -\frac{126 \text{Li}_6\left(\frac{1}{2}\right)}{13}-\frac{162 \text{Li}_6\left(-\frac{1}{2}\right)}{13}+\frac{\text{Li}_6\left(-\frac{1}{8}\right)}{39}+\frac{3 \zeta (3)^2}{8}+\frac{31}{16} \zeta (5) \log (2)-\frac{4783 \pi ^6}{1179360}-\frac{1}{208} \log ^6(2)+\frac{1}{208} \pi ^2 \log ^4(2)-\frac{1}{156} \pi ^4 \log ^2(2)$$ I believe this is insurmountable via your adored "elementary/artistic approaches". Similar thing could happen for the triple sum $S$ given above, but so far nothing is certain given my limited knowledge. $\endgroup$
    – pisco
    Commented Aug 29, 2020 at 4:04
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    $\begingroup$ @AliShather Yes, I am confident on its correctness. Unfortunately, I am afraid your eagerness is not in tandem with your capability, you might need a decade of learning in order to understand the paper. Even I am not (yet) able to fully understand everything there. The expression for $A$ was written as a conjecture there, but an author told me it was subsequently proved, the proof will be included in a revision. $\endgroup$
    – pisco
    Commented Aug 29, 2020 at 4:19

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