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I am wondering if there is a closed form for

$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n},$$

Where $a$ is positive real.

The integral representation of this sum is

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

I encountered this sum while trying to generalize $\displaystyle\int_0^1 \frac{\ln \left( 1+x^{2+\sqrt{3}}\right)}{1+x}dx$ to $\displaystyle\int_0^1 \frac{\ln \left( 1+x^{a}\right)}{1+x}dx.$

Thanks,

Ali Shadhar
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    Have you already checked here? It's admittedly not quite what you want, since you want a slight modification of $\mathbf{H}_{1}^{(1)}(-1)$, but the techniques there are quite interesting for problems of this kind. – Brevan Ellefsen Nov 24 '21 at 07:48

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A proof for $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n} =\frac{5\pi^2}{48}-\frac14 \ln^22$$ can be found at Evaluating $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$, and in a comment there Omram Kouba provides the general result

$$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_{an}}{n}$$

via link to their paper: http://arxiv.org/abs/1010.1842. Specifically, it is there shown that: $$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_{an}}{n} = \frac{(a^2+1)\pi^2}{24a} - \frac{1}{2} \sum_{j=0}^{a-1} \log^2\left(2 \sin \frac{(2j+1)\pi}{2a} \right)$$

Brevan Ellefsen
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