Csdrhrt

Apologies if this is not mathematically very precise.

I have been trying to calculate the Fourier series of $e^{i q |m|}$, but I'm having trouble with the absolute value in the exponential.
Without having a proof, I think this might be true, but I'm not sure.
$$sum_{m=-infty}^infty e^{i(q|m|-km)}=1+sum_{m=1}^inftyleft(e^{i m(q-k)}+e^{i m(q+k)}right)
=1+pi(delta(q-k)+delta(q+k))$$
Splitting up $e^{i q|m|}=cos(qm) + i |sin(qm)|$
the second term in the above formula would correspond to the cosine part, but it seems wrong that the sine part becomes just 1?

--------- EDIT -------

I now believe that the LHS equals just $pi(delta(q-k)+delta(q+k))$ (I've done the sum from -5000 to 5000 and plotted the real and imaginary parts). I still don't know how to show this though.

--------- EDIT2 --------

Actually, it seems that there is also some part like $idelta'(k-q)+idelta'(k+q)$ or something similar.

We can make some progress by taking the real part, which is straightforward and then calculate the imaginary part via the Kramers-Kronig relations. The latter requires the function in question to be analytic in the UHP and vanish as $|k|toinfty$.

This is why we'll consider the complex conjugate instead
$$chi(k)=sum_{m=-infty}^infty e^{i(km-q|m|)}.$$
I'm not entirely sure if the function is analytic. If it isn't, that might explain the discrepancy of this calculation to the actual solution.

The real part is
$$Re[chi(k)]=1+sum_{m=1}^inftyleft[cos((q-k)m)+cos((q+k)m)right]
=pileft[delta(k-q)+delta(k+q)right],$$
such that the imaginary part evaluates to
$$Im[chi(k)]=-frac1pimathcal Pint_{-infty}^inftyfrac{Re[chi(k')]}{k'-k},dk'
=frac{1}{k-q}+frac{1}{k+q}.$$

However, this answer is not quite right, as can be seen from the numerical solution. The actual answer should be
$$Im[chi(k)]=frac{1}{k-q}-frac{1}{k+q},$$
the crucial difference being the minus sign, which is probably related to the absolute value being nonanalytical at 0.

Blue=actual answer, grey=numerically performed sum (oscillatory), green=average of grey.

Another way to write the solution is
$$sum_{m=-infty}^infty e^{ikam-ik_0a|m|}
=frac ialeft(frac 1{k-k_0+ivarepsilon}-frac 1{k+k_0-ivarepsilon}right)$$
(understood in the usual sense that you have to take the limit $varepsilonto0^+$ in the end).