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I am trying to evaluate

$$I=\int_0^1\frac{\ln(1+x^2)\text{Li}_2(x)}{x}dx$$

Integration by parts yields

$$I=\frac58\zeta(4)-\frac12\int_0^1\frac{\ln(1-x)\text{Li}_2(-x^2)}{x}dx$$

Another related integral is

$$\int_0^1 \frac{x\ln^2x\ln(1-x)}{1+x^2}dx$$

Any idea now to evaluate any of the integrals above without using the generating function $\sum_{n=1}^\infty\frac{H_n}{n^3}x^n$ ?

Thanks

Ali Shadhar
  • 25,498
  • Sooner or later we have to invoke Euler sums, but maybe a viable approach is to compute the FL-expansion of $\log(1+x^2)/x$ and combine it with the FL-expansion of $\text{Li}_2$ (https://math.stackexchange.com/a/3731241/44121) – Jack D'Aurizio Jun 23 '20 at 15:47
  • Anyway, as a starting point for other users, $\int_{0}^{1}x^{2n-1}\text{Li}2(x),dx=\frac{\pi^2}{12n}-\frac{H{2n}}{4n^2} $, so our integral equals $$ \sum_{n\geq 1}\frac{(-1)^n H_{2n}}{4n^3}-\sum_{n\geq 1}\frac{(-1)^n \pi^2}{12n^2}=\color{red}{\sum_{n\geq 1}\frac{(-1)^n H_{2n}}{4n^3}}-\frac{\pi^4}{144} $$ – Jack D'Aurizio Jun 23 '20 at 16:00
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    Thanks @Jack D'Aurizio . The point of my question is to find a different solution for $\Re\text{Li}_4(1+i)$. – Ali Shadhar Jun 23 '20 at 20:46

0 Answers0