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this integral got posted on a mathematics group by a friend $$I=\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$$ I tried seeing what I'd get from integrating by parts which got me $$I=\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\arcsin ^2\left(x\right)}{1-x}\:\mathrm{d}x-\int _0^1\frac{\ln ^2\left(x\right)\arcsin \left(x\right)\ln \left(1-x\right)}{\sqrt{1-x^2}}\:\mathrm{d}x$$ and these seem quite hard to evaluate, I also tried expanding the integral in series but got nowhere.

I also tried using expressing $I$ the following way $$I=\frac{1}{2}\int _0^1\frac{\arcsin ^2\left(x\right)\ln ^2\left(x\right)}{x}\:\mathrm{d}x+\int _0^1\frac{\arcsin ^2\left(x\right)\ln ^2\left(1-x\right)}{x}\:\mathrm{d}x$$ $$-\frac{1}{2}\int _0^1\frac{\arcsin ^2\left(x\right)\ln ^2\left(\frac{x}{1-x}\right)}{x}\:\mathrm{d}x$$ and the first $2$ integrals are probably doable but I can't think of a way to approach the last one.

Is it possible to evaluate $I$ or the third integral on the last equation in a fairly simple way?

Edit: By Dilogarithm functional equations the following are also related $$\int _0^1\frac{\arcsin ^2\left(x\right)\operatorname{Li}_2\left(x\right)}{x}\:\mathrm{d}x,\:\int _0^1\frac{\arcsin ^2\left(x\right)\operatorname{Li}_2\left(1-x\right)}{x}\:\mathrm{d}x$$ and therefore some very rough hamornic series are involved.

logandetner
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    In your last question, you asked a series without knowing its result, the result turns out to contain polylog constants. The result for this one is even more unfriendly. – pisco Apr 26 '23 at 07:00
  • Well I'm aware that this integral essentially is related to harmonic series of weight 5 structure so we'll be having a closed form in terms of classical linear elements like $\ln \left(2\right)\zeta \left(4\right)$, $\ln \left(2\right)\operatorname{Li}_4\left(\frac{1}{2}\right)$ ,$\operatorname{Li}_5\left(\frac{1}{2}\right)$ though I'm suspicious it could contain other elements given it is not a linear one in nature. – logandetner Apr 26 '23 at 07:18
  • It's indeed weight $5$, but it has level 4, what you said only applies for level 2. In this weight and level, the space (dimension 32) has 3 ugly constants which is (presumably) not a simple combination of polylog. One such ugly constant appear in the result. – pisco Apr 26 '23 at 07:21
  • Very interesting! perhaps some of them involve $\operatorname{Im} \text{Li}_5\left(\frac{1+i}{2}\right),:\operatorname{Im} \text{Li}_5\left(1+i\right)$? you could post its closed form if you want, I wonder if it can be evaluated in a simple and tricky way. – logandetner Apr 26 '23 at 07:26
  • One of the integrals in the post also contain $\operatorname{Re} \text{Li}_5\left(1+i\right)$ which I think can be expressed in terms of known constants. – logandetner Apr 26 '23 at 07:34
  • $\operatorname{Im} \text{Li}_5\left(\frac{1+i}{2}\right),:\operatorname{Im} \text{Li}_5\left(1+i\right)$ are related, and they're not those 3 ugly constants, neither is $\operatorname{Re} \text{Li}_5\left(1+i\right)$, it is of level 2, so again has a pretty formula in terms of $\text{Li}_5(1/2)$ etc. – pisco Apr 26 '23 at 07:39
  • If it can help, the integral is equal to:$$I=\sum_{k=1}^{\infty}\frac{1}{k}\int_0^\infty te^{-kt}\arcsin^2(e^{-t})dt$$ – Zima Apr 26 '23 at 08:40

1 Answers1

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$$I=\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x^{2}\right)}{x}\:\mathrm{d}x==\zeta_2\zeta_3 +\zeta_2\ln^3 2 -\frac{31}{32}\zeta_5-\frac{15}{8}\zeta_4\ln2$$

user178256
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  • May you share how exactly you got this answer? – Accelerator Apr 27 '23 at 22:07
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    Note that this integral is different than the one proposed in the OP, your integral $I$ is: $\displaystyle \int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x^2\right)}{x}:dx=\frac{1}{8}\sum _{k=1}^{\infty }\frac{4^kH_k}{k^4\binom{2k}{k}}+\frac{1}{8}\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^3\binom{2k}{k}}-\frac{1}{8}\zeta \left(2\right)\sum _{k=1}^{\infty }\frac{4^k}{k^3\binom{2k}{k}}$, and you may find those series evaluated here. – Jorge Layja Apr 28 '23 at 02:10