I'm trying to find a handy way to find the infinite sum
$ \sum_{n=1}^\infty \frac{cos( a n)}{n^2} $
through a Fourier series. The regular sum can be evaluated using Mathematica fairly easily, and gives a result of $ \sum_{n=1}^\infty \frac{cos( a n)}{n^2} = 0.324138$, found through some Polylogs.
The idea would be to FT $\frac{cos( a n)}{n^2}$ and evaluate the sum in reciprocal space but this isn't yielding anything good. I can get the FT no problem but when I sum over the reciprocal lattice vectors the sum diverges. I'm using $ \vec{a} = l \hat{x} $ for real space vectors, where $l=1$ is the separation of sampling sites, and $\vec{b} = \frac{1}{l} \hat{x} $ for reciprocal space vectors. It's fairly easy to see from the Fourier Transform F from real space to reciprocal space ($t$ - bad choice I know) that
$ {F} (\frac{cos( a n)}{n^2}) \to \frac{1}{2} \sqrt{\pi/2} (Sign[1 - t] + t Sign[-1 + t] + Sign[1 + t] + t Sign[1 + t]) $
so will diverge when doing the sum over t (which goes as 1/n right?), as this Fourier Transformed sum now goes as the harmonic series.
Can anyone see where I've slipped up in my reasoning? Cheers.
This can be evaluated using the polylogarithm function $\mathrm{Li}_2$: $$ \begin{align} \sum_{n=1}^\infty\frac{\cos(an)}{n^2} &=\mathrm{Re}\left(\sum_{n=1}^\infty\frac{e^{ian}}{n^2}\right)\\ &=\mathrm{Re}\left(\mathrm{Li}_2(e^{ia})\right)\\[2pt] &=\frac12\left(\mathrm{Li}_2(e^{ia})+\mathrm{Li}_2(e^{-ia})\right)\tag{1} \end{align} $$ Note that the derivative of $(1)$ is $$ \begin{align} -\sum_{n=1}^\infty\frac{\sin(an)}{n} &=-\mathrm{Im}\left(\sum_{n=1}^\infty\frac{e^{ian}}{n}\right)\\ &=\mathrm{Im}\left(\log\left(1-e^{ia}\right)\right)\\[4pt] &=\frac{a-\pi}{2}\quad\text{on }(0,2\pi)\tag{2} \end{align} $$ Integrating $(2)$ and noting that $(1)$ is $-\frac{\pi^2}{12}$ at $a=\pi$ yields that on $(0,2\pi)$, $$ \sum_{n=1}^\infty\frac{\cos(an)}{n^2} =\left(\frac{a-\pi}{2}\right)^2-\frac{\pi^2}{12}\tag{3} $$
Plot of one period:
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