Quadratic Euler Sum $\sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}\left(\sum_{m=1}^n \frac{(-1)^{m-1}}{2m-1}\right)^2$

Question

Question: How to evaluate $$\sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}\left(\sum_{m=1}^n \frac{(-1)^{m-1}}{2m-1}\right)^2$$ I have computed a class of polylog integrals of considerable complexity, but this sum is hardly related to any of them in a direct sence. Any help or suggestions will be appreciated.

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Update: See arXiv $2007.03957$ for a solution utilizing the power of MZVs.

Answer 1

This answer provides a list of polylog integrals to be solved which can be checked with your list of 172 polylog integrals.

Let us consider the generating function

$$g(z) = \sum_{n=1}^{\infty} z^n J_{n}^2\tag{1}$$

from which we can derive the sums in question by integrating with respect to $z$ and taking the values at $z=\pm1$.

As for instance

$$s_1 = \sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}J_n^2=\int_0^1 \log \left(\frac{1}{z}\right)\frac{ g(t z)}{t z} \, dz\;\; |_{t\to \pm 1}\tag{2}$$

which follows from

$$\int_0^1 \log \left(\frac{1}{z}\right) (t z)^{n-1} \, dz = \frac{t^{n-1}}{n^2}|_{t\to \pm 1}= \frac{(\pm 1)^{n-1}}{n^2}\tag{2a}$$

An integral representation of $J$ is found by writing

$$J(n) = \sum _{k=1}^n (-1)^{k+1}\frac{1}{2 k-1} = \sum _{k=1}^{n} (-1)^{k+1}\int_0^1 x^{2k-2}\,dx \\ =\int_0^1 \sum _{k=1}^n (-1)^{k+1} x^{2 k-2}\,dx = \int_0^1 \left(-\frac{(-1)^n x^{2 n}-1}{x^2+1}\right)\,dx \tag{3}$$

Hence we can write $g$ as a double integral

$$g(z) = \int_{[0,1]^2} \sum _{n=1}^{\infty } \frac{\left(-\left((-1)^n x^{2 n}-1\right)\right) \left(-\left((-1)^n y^{2 n}-1\right)\right) z^{n-1}}{\left(x^2+1\right) \left(y^2+1\right)}\,dx\,dy=\int_{[0,1]^2} \frac{1-x^2 y^2 z^2}{(1-z) \left(x^2 z+1\right) \left(y^2 z+1\right) \left(1-x^2 y^2 z\right)}\,dx\,dy\tag{4}$$

The integrals can be done one by one leading finally to the rather lengthy result

$$g(z) = \frac{1}{96 (1-z) \sqrt{z}}\left(\pi ^2+24 \operatorname{Li}_2\left(\frac{i+\sqrt{z}}{i-\sqrt{z}}\right)+12 \operatorname{Li}_2\left(-\frac{\left(\sqrt{z}-1\right)^2}{\left(\sqrt{z}+1\right)^2}\right)+24 \operatorname{Li}_2\left(\frac{i-\sqrt{z}}{i+\sqrt{z}}\right)-24 \operatorname{Li}_2\left(\frac{-\sqrt{z} (1-i)-i+z}{\sqrt{z} (1-i)-i+z}\right)-24 \operatorname{Li}_2\left(\frac{i-(1+i) \sqrt{z}+z}{i+(1+i) \sqrt{z}+z}\right)-24 \pi \sin ^{-1}\left(\sqrt{\frac{z}{z+1}}\right)+24 \pi \tan ^{-1}\left(\sqrt{z}\right)-48 i \log \left(\frac{(2+2 i) \left(z-i \sqrt{z}\right)}{\sqrt{z}+1}\right) \sin ^{-1}\left(\sqrt{\frac{z}{z+1}}\right)+48 i \log \left(\frac{(2-2 i) \left(z+i \sqrt{z}\right)}{\sqrt{z}+1}\right) \sin ^{-1}\left(\sqrt{\frac{z}{z+1}}\right)+48 i \log \left(\sqrt{z}-i\right) \sin ^{-1}\left(\sqrt{\frac{z}{z+1}}\right)-48 i \log \left(\sqrt{z}+i\right) \sin ^{-1}\left(\sqrt{\frac{z}{z+1}}\right)+48 \log \left(\frac{(2+2 i) \left(z-i \sqrt{z}\right)}{\sqrt{z}+1}\right) \tanh ^{-1}\left(\sqrt{z}\right)+48 \log \left(\frac{(2-2 i) \left(z+i \sqrt{z}\right)}{\sqrt{z}+1}\right) \tanh ^{-1}\left(\sqrt{z}\right)-48 \log \left(\frac{2 (z+1)}{\left(\sqrt{z}+1\right)^2}\right) \tanh ^{-1}\left(\sqrt{z}\right)-96 \log (z+1) \tanh ^{-1}\left(\sqrt{z}\right)\right)\tag{5}$$

For the convenience of the reader here is the Mathematica expression:

-(1/(96 (-1 + z) Sqrt[
  z]))(\[Pi]^2 - 24 \[Pi] ArcSin[Sqrt[z/(1 + z)]] + 
   24 \[Pi] ArcTan[Sqrt[z]] + 
   48 I ArcSin[Sqrt[z/(1 + z)]] Log[-I + Sqrt[z]] - 
   48 I ArcSin[Sqrt[z/(1 + z)]] Log[I + Sqrt[z]] - 
   96 ArcTanh[Sqrt[z]] Log[1 + z] - 
   48 ArcTanh[Sqrt[z]] Log[(2 (1 + z))/(1 + Sqrt[z])^2] - 
   48 I ArcSin[Sqrt[z/(1 + z)]] Log[((2 + 2 I) (-I Sqrt[z] + z))/(
     1 + Sqrt[z])] + 
   48 ArcTanh[Sqrt[z]] Log[((2 + 2 I) (-I Sqrt[z] + z))/(
     1 + Sqrt[z])] + 
   48 I ArcSin[Sqrt[z/(1 + z)]] Log[((2 - 2 I) (I Sqrt[z] + z))/(
     1 + Sqrt[z])] + 
   48 ArcTanh[Sqrt[z]] Log[((2 - 2 I) (I Sqrt[z] + z))/(
     1 + Sqrt[z])] + 24 PolyLog[2, (I - Sqrt[z])/(I + Sqrt[z])] + 
   24 PolyLog[2, (I + Sqrt[z])/(I - Sqrt[z])] + 
   12 PolyLog[2, -((-1 + Sqrt[z])^2/(1 + Sqrt[z])^2)] - 
   24 PolyLog[
     2, (-I - (1 - I) Sqrt[z] + z)/(-I + (1 - I) Sqrt[z] + z)] - 
   24 PolyLog[2, (I - (1 + I) Sqrt[z] + z)/(I + (1 + I) Sqrt[z] + z)])

The function $g(z)$ is real and looks like this

From $(5)$ and $(2)$ you can identify the polylog integrals to be solved.

Discussion

Close to $z=\pm1$ we have approximately

$$g(z\sim -1) \simeq \frac{1}{8} \pi \log \left(\frac{2}{z+1}\right)$$

$$g(z\sim +1) \simeq \frac{\pi ^2}{32}-\frac{\pi ^2}{16 (z-1)} $$