3

I was evaluating the integral $$\mathcal{I}=\int_0^1 \frac{\ln^2 x\ln(1-x)}{x^2+1}\mathrm dx$$

I tried to use the series $$\ln(1-x)=-\sum_{n=1}^\infty \frac{x^n}{n}$$ and interchanged the sum and the integral. $$\mathcal{I}=-\sum_{n=1}^\infty \frac{1}{n}\overbrace{\int_0^1 \frac{x^n\cdot \ln^2 x}{x^2+1}\mathrm dx}^{\mathcal{J}}$$ Now, it's quite obvious that $$\mathcal{J}=\frac{\mathrm d^2}{\mathrm dn^2}\int_0^1 \frac{x^n}{x^2+1}\mathrm dx$$ Now, I found in the answer here by @Olivier Oloa that $$\int_0^1 \frac{x^m}{1+x^n}\mathrm dx=\frac{1}{2n}\psi\bigg(\frac{m+n+1}{2n}\bigg)-\frac{1}{2n}\psi\bigg(\frac{m+1}{2n}\bigg)$$. So, \begin{align}\mathcal{J}&=\frac{\mathrm d^2}{dn^2}\bigg[\frac{1}{4}\psi\bigg(\frac{n+3}{4}\bigg)-\frac{1}{4}\psi\bigg(\frac{n+1}{4}\bigg)\bigg]\\&=\frac{1}{64}\bigg(\psi^{(2)}\bigg(\frac{n+3}{4}\bigg)-\psi^{(2)}\bigg(\frac{n+1}{4}\bigg)\bigg)\end{align} So, the integral converts to the scary-looking series $$\boxed{\mathcal{I}=\frac{-1}{64}\sum_{n=1}^\infty \frac{1}{n}\bigg(\psi^{(2)}\bigg(\frac{n+3}{4}\bigg)-\psi^{(2)}\bigg(\frac{n+1}{4}\bigg)\bigg)}$$ Can someone hint on how to evaluate this infinite series? I want a solution that is based on this approach that I gave.

  • 2
    I don't think that converting this into a series is a good idea. I might post a solution. – Laxmi Narayan Bhandari Jul 11 '21 at 07:01
  • 1
    Your question is a duplicate, you may find ways to go here, I suggest you make use of Approach Zero. – Jorge Layja Jul 11 '21 at 07:20
  • 2
    @JorgeLayja This is the first time that Approachzero couldn't find that this is already asked. Anyways, thank you. I will, therefore, close this question. –  Jul 11 '21 at 07:22
  • I still suggest you remove this question so to prevent repeated questions flooding this site, you can post any alternative solution you come up with there. – Jorge Layja Jul 11 '21 at 07:25
  • 1
    I want an answer based on my approach. –  Jul 11 '21 at 07:25
  • @JorgeLayja This solution is not complete. I still want some hints from you on how to evaluate this series. –  Jul 11 '21 at 07:47
  • If a step in my attempt is wrong, you can always leave a comment. –  Jul 11 '21 at 08:40
  • 1
    With Mathematica and MZIntegrate packages I have: $$\frac{\pi ^2 C}{3}-\frac{\zeta \left(4,\frac{1}{4}\right)}{64}+\frac{\zeta \left(4,\frac{3}{4}\right)}{64}+\frac{1}{32} \pi ^3 \ln (2)\approx -0.27075$$ where $C$ is Catalan const. – Mariusz Iwaniuk Jul 11 '21 at 10:17
  • 3
    It is hard to fix the question in the OP. We have an integral $\mathcal I$, which can be computed almost algorithmically, but instead of doing this we complicate maximally the computation of that integral, the digamma function $\psi$ appears, and in a final formula we have expressed $\mathcal I$ as a series involving terms which involve the second derivative of $\psi$ computed in rational, non-integral points. In bold face we have an imperative request. If a potential answerer is not allowed to go back to the definition of $\mathcal I$, which formula for $\psi''$ should be used? – dan_fulea Jul 11 '21 at 13:18

0 Answers0