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How can I calculate the integral? Any ideas ?

$$\int_0^1 \frac{\ln y\operatorname{Li}_2 (-y)}{1-y^2} \, dy$$

I thought about using this formula but I can not get to something, someone can prove this? $$\int_0^1 \frac{\ln^{s-1} x\operatorname{Li}_2(x)}{1-x}dx = (-1)^{s-1}(s-1)!\frac{\zeta(s)-\zeta(2s)} 2$$

J. M. ain't a mathematician
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whitexlotus
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  • Note that $$ \frac{\log (x) \operatorname{Li}_2(-x)}{1-x^2} \simeq-\frac{1}{(x-1)^3} \quad \text{ for } x\to 1. $$ Hence the integrand is not integrable, around 1. – Idividedbyzero Jun 24 '25 at 11:17

1 Answers1

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$$I=\int_0^1 \frac{\ln(x)\operatorname{Li}_2(-x)}{1-x^2}\, dx$$

I applied partial fractions,

$$I\implies \frac12\int_0^1 \frac{\ln(x)\operatorname{Li}_2(-x)}{1+x}\, dx+\frac12\int_0^1 \frac{\operatorname{Li}_2(-x)\ln(x)}{1-x}\, dx$$

Using the below formula from here,

$$\color{red}{\sum_{n=1}^\infty H_n^{(p)}x^n=\frac{\operatorname{Li}_p(x)}{1-x}\underset{\text{p=2, x=-x}}\implies \sum_{n=1}^\infty (-1)^nH_n^{(2)}x^n=\frac{\operatorname{Li}_2(-x)\ln(x)}{1+x}}$$

We have these,

$$I\implies \frac12\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1 x^n \ln(x)\, dx+\frac12\int_0^1 \frac{\operatorname{Li}_2(-x)\ln(x)}{1-x}\, dx$$

Now umm for this second integral, we can stick to using the definition

$$I\implies \frac12\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1 x^n \ln(x)\, dx+\frac12\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 \frac{x^n\ln(x)}{1-x}\, dx$$

We know that $\int_0^1 \frac{x^n \ln(x)}{1 - x} \, dx = H_n^{(2)} - \zeta(2)$

$$I\implies \frac12\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1 x^n \ln(x)\, dx+\frac12\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 \frac{x^n\ln(x)}{1-x}\, dx$$

$$I\implies \frac{1}{2} \sum_{n=1}^{\infty} (-1)^n \left( \frac{1}{n^2} - H_n^{(2)} \right) \int_0^1 x^{n-1} \ln(x) \, dx + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \left( H_n^{(2)} - \zeta (2) \right)$$

$$I\implies -\frac{1}{2} \sum_{n=1}^{\infty} (-1)^n \left( \frac{1}{n^2} - H_n^{(2)} \right) \left( \frac{1}{n^2} \right) + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \left( H_n^{(2)} - \zeta (2) \right)$$

Using the polylogarithm definition,

$$\implies \frac{1}{2} \sum_{n=1}^{\infty}(-1)^n \frac{H_n^{(2)}}{n^2} - \frac{1}{2} \operatorname{Li}_4 (-1) + \frac{1}{2} \sum_{n=1}^{\infty} (-1)^n\frac{H_n^{(2)}}{n^2} - \frac{1}{2} \zeta (2) \operatorname{Li}_2 (-1)$$

Using standard values I took from WA,

$$\implies \frac{17}{1440} \pi^4+\sum_{n=1}^{\infty} (-1)^n\frac{H_n^{(2)} }{n^2}$$

See here,

$$\sum_{n=1}^{\infty}(-1)^n\frac{H_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln(2)\zeta(3)+\ln^2(2)\zeta(2)-\frac16\ln^4(2)$$

So that completes our integral,

$$\int_0^1 \frac{\ln(x)\operatorname{Li}_2(-x)}{1-x^2}\, dx =\frac{17}{1440}\pi^4-4\operatorname{Li}_4 \left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln(2)\zeta(3)+\ln^2(2)\zeta(2)-\frac16\ln^4(2)$$

You can get numerical checks here and here

Amrut Ayan
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