# Compute infinite sum of modified Bessel functions and Cos

I have encountered the expression $$\sum_{n >0} K_0(2 \pi \rho n) \cos(2\pi n z)\, ,$$ for which I would like to obtain an explicit, closed expression (that means for me getting rid of any infinite sum). $K_0$ is the $0^{\rm th}$ modified Bessel function of second kind, $\rho \in \mathbb{R}^+$ and $z \in \mathbb R$. I didn't find a closed expression for this quantity in the literature up to now. Does anybody have a clue how to simplify this expression?

• $K_0$ decays so fast that such series is approximately equal to its first term $K_0(2\pi\rho)\cos(2\pi z)$. – Jack D'Aurizio Mar 12 '18 at 16:57
• The literature about Maass forms may be useful. – graveolensa Mar 13 '18 at 22:08

$$K_0(2\pi \rho n)=\int_{0}^{+\infty}\frac{\cos(2\pi\rho n t)}{\sqrt{t^2+1}}\,dt=\int_{0}^{+\infty}\frac{\cos(2\pi nt)}{\sqrt{t^2+\rho^2}}\,dt=\int_{0}^{+\infty}\frac{t\sin(2\pi nt)}{2\pi n(t^2+\rho^2)^{3/2}}\,dt$$ hence by multiplying both sides by $\cos(2\pi nz)$ and summing over $n\geq 1$, by the Fourier series of the sawtooth wave
$$\sum_{n\geq 1}\frac{\sin(2\pi n\theta)}{2\pi n}=\frac{1}{4}-\frac{\{\theta\}}{2}$$
we get $$\sum_{n\geq 1}K_0(2\pi \rho n)\cos(2\pi n z)=\int_{0}^{+\infty}\frac{t}{(t^2+\rho^2)^{3/2}}\cdot\frac{\{t-z\}-\{t+z\}}{4}\,dt$$ hence the LHS equals zero if $z\in\frac{1}{2}\mathbb{Z}$ and it is bounded by $\frac{1}{2\rho}$ in absolute value, for any $\rho>0$.
• Here $\{ x\}$ stands for the fractional part of $x$, i.e. $x-\lfloor x\rfloor$. – Jack D'Aurizio Mar 12 '18 at 17:29