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Question. I was given that $$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2=\frac{32}\pi G\ln2+\frac{64}\pi\Im\operatorname{Li}_3\left(\frac{1+i}2\right)-2\ln^22-\frac53\pi^2$$ where $H_n$ harmonic numbers, $G$ Catalan and $\operatorname{Li}_n$ polylogarithm. How can it be proved?

My Approach. Using $\int_0^1x^{n-1}\ln(1-x)dx=-\frac{H_n}n$ one have $$S=\int_0^1-\ln(1-x)\sum_{n=1}^\infty\left(\frac{(2n)!}{4^n(n!)^2}\right)^2x^{n-1}dx =\int_0^1-\ln(1-x)\left(\frac2\pi\frac{\mathbf{K}(x)}x-\frac1x\right)dx$$ where $\mathbf{K}$ denotes elliptic integral of the first kind. The question boils down to finding $$\int_0^1\frac{\mathbf{K}(x)\ln(1-x)}xdx$$ For this integral, I tried to use the integral representation of the elliptic integral and got: $$\int_{(0,1)^3}\frac{dxdydz}{\sqrt{1-y^2}\sqrt{1-xy^2}(zx-1)}$$ This is the furthermost step I can get.

Infiniticism
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Kemono Chen
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    Perhaps it was a typo, but Catalans constant is defined as $$\mathrm G=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$$ – clathratus May 10 '19 at 16:13
  • Sorry for that. I have corrected it. – Kemono Chen May 10 '19 at 16:20
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    Although I'm almost certain I have encountered this one before (but my mind might been tricked) I wasn't able to find it again. However, while searching I stumbled upon lots and lots of related sums (together with proofs for their values!) and would like to share them. Especially the work of John M. Campbell is quite fruitful, see for example here and here. Moreover this website contains some very interesting sums (also closey related!). – mrtaurho May 10 '19 at 19:43
  • I start trying to prove, by using the second link of mrtaurho starting with equation 2.1 and the identity $$\text{Hypergeometric2F1}\left[\frac{3}{2},\frac{3}{2},3,z\right]==\frac{16 (-2 \text{EllipticE}[z]+(2-z) \text{EllipticK}[z])}{\pi z^2}$$. – stocha May 11 '19 at 10:50
  • who gave you this series/where did you find it? – clathratus May 11 '19 at 22:04
  • @clathratus I found that in a discussion group of definite integrals of my country. – Kemono Chen May 12 '19 at 03:22
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    BTW: \begin{align}\int_0^1x^{n-1}\ln(1-x)dx=-\frac{H_n}n,\end{align} – FDP May 12 '19 at 19:41

1 Answers1

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One have FL expansion $$\small K(x)=\sum _{n=0}^{\infty } \frac{2 P_n(2 x-1)}{2 n+1},\ \ \frac{\log (1-x)}{x}=\sum _{n=0}^{\infty } 2 (-1)^{n-1} (2 n+1) P_n(2 x-1) \left(\sum _{k=n+1}^{\infty } \frac{(-1)^{k-1}}{k^2}\right)$$ Thus by orthogonal relation of Legendre polynomial and series-integral conversion one have: $$\int_0^1 \frac{K(x) \log (1-x)}{x} \, dx=16 \int_0^1 \frac{\log (x) \tanh ^{-1}(x)}{x^2+1} \, dx$$ RHS has a polylogarithmic primitive (up to trilog) hence trivial.

Infiniticism
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