# Mellin transform of a general term of an alternating series

What is the Mellin transform of the following function $$f(n)=\frac{(-1)^{n-1}}{n^2}\cos an \ ?$$

I encountered this problem while evaluating the series $$\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}\cos an$$. To evaluate this series by Mellin summation formula, we need to find the Mellin transform of the general term at first. Without the $$(-1)^{n-1}$$ term, the transform is a piece of cake. But for the alternating sign, I could not find a suitable transformation formula for this $$f(n)$$. Is there any general formula for the type $$(-1)^nf(n)$$ in Mellin transform? Any help is appreciated.

Observe that $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} \cos(a n) \\ \quad = \sum_{n=1}^\infty \frac{1}{n^2} \cos(a n) - 2\sum_{n=1}^\infty \frac{1}{(2n)^2} \cos(a (2n)) \text{,}$$ if the two series on the right converge. (That is, ignore the signs first, then subtract off two copies of the even indexed terms, so that those terms acquire the correct signs.)
The Mellin transform of $$\frac{1}{n^2} \cos(a n)$$ is $$-a^{2-s}\Gamma(s-2) \cos(\pi s/2)$$ and of $$(-2)\frac{1}{(2n)^2} \cos(a (2n))$$ is $$2^{1-s}a^{2-s}\Gamma(s-2) \cos(\pi s/2)$$. (The inverse Mellin transform of the sum of these recovers $$f$$, which is a good sign.)
This may not be of much use to you. The sum you are evaluating is $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} \cos(a n) = \frac{-1}{2} \left(\text{Li}_2\left(-\mathrm{e}^{-\mathrm{i} a}\right) + \text{Li}_2\left(-\mathrm{e}^{\mathrm{i} a}\right)\right) \text{,}$$
where $$\mathrm{Li}_2$$ is the dilogarithm, which does not appear in many Mellin transform tables.