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Prove $$\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2} \sum_{j=0}^{\infty} \frac{{(-1)}^j}{2k+j+1}=-\frac{\pi^2}{12}\ln{2}+\pi C-\frac{33}{16} \zeta(3)$$ where C is catalan's constant. Wolfram Alpha confirms that the sums converge to approximately the right side. Wolfram Alpha also evaluates the first sum in terms of Hurwitz lerch transcendent or digamma function but how do I then evaluate the outer sum with either of these functions.

Original question is $$\int_0^1 \frac{\text{Li}_2(-x^2)}{1+x} \; \mathrm{d}x$$ and I've got it to the double sum here by writing Li as its series form and forming a geometric series with $\frac{1}{1+x}$.

Any tips or suggestions? maybe other approach to the integral?

Edit: Integration by parts may work better? $$\ln{(1+x)}\text{Li}_2(-x^2) \bigg \rvert_0^1 + 2\int_0^1 \frac{\ln{(1+x)}\ln{(1+x^2)}}{x} \; \mathrm{d}x$$ Wolfram says that second integral is $\pi C -\frac{33 \zeta(3)}{16}$ which is very good here but I don't know how to evaluate that integral.

$$\int_0^1 \frac{2\ln{(1+x)}\ln{(1+x^2)}}{x} \; \mathrm{d}x=\int_0^1 \frac{\ln^2{(1+x)(1+x^2)}}{x} \; \mathrm{d}x-\int_0^1 \frac{\ln^2{(1+x)}}{x} \; \mathrm{d}x - \int_0^1 \frac{\ln^2{(1+x^2)}}{x} \; \mathrm{d}x$$ Last integral is 0 $$\int_0^1 \frac{2\ln{(1+x)}\ln{(1+x^2)}}{x} \; \mathrm{d}x=\int_0^1 \frac{\ln^2{(1+x)(1+x^2)}}{x} \; \mathrm{d}x-\int_0^1 \frac{\ln^2{(1+x)}}{x} \; \mathrm{d}x$$

  • I believe integrals such as $\int_0^1 \frac{\text{Li}_2(-x^2)}{1+x} ; \mathrm{d}x$ has been completed tabulated in https://arxiv.org/abs/1911.12155 due to @User628759. Methods used there are quite systematic and nice. Its appendix also contains a compendium of Euler sum (page 111), which should include your case. – pisco Aug 04 '20 at 14:34
  • Have a look at the 3rd formula stated in this question https://math.stackexchange.com/questions/3294203/is-there-a-closed-form-solution-for-sum-n-1-infty-sum-m-1-infty-fr – Donald Splutterwit Aug 04 '20 at 14:50

1 Answers1

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We have that

$$\sum_{j=0}^{\infty} \frac{{(-1)}^j}{2k+j+1}=\sum_{j=1}^{\infty} \frac{{(-1)}^{j+1}}{j}-\sum_{j=1}^{2k} \frac{{(-1)}^{j+1}}{j}=\ln 2-\sum_{j=1}^{2k} \frac{{(-1)}^{j+1}}{j}$$

then

$$\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2} \sum_{j=0}^{\infty} \frac{{(-1)}^j}{2k+j+1} =\ln 2\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2}-\sum_{k=1}^{\infty} \frac{{(-1)}^n}{k^2}\sum_{j=1}^{2k} \frac{{(-1)}^{j+1}}{j}$$

with

$$\ln 2\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2}=-\frac{\pi^2}{12}\ln{2}$$

and using the results indicated here and here

$$\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2}\sum_{j=1}^{2k} \frac{{(-1)}^{j+1}}{j}=\sum_{k=1}^{\infty} \frac{{(-1)}^kH_{2k}}{k^2}-\sum_{k=1}^{\infty} \frac{{(-1)}^k H_k}{k^2}= \frac{23}{16} \zeta (3) - \pi \mathbf{G}+\frac{5}{8}\zeta(3)$$

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