Series $\sum_{n=1}^{\infty} \left[K_0\left(\sqrt{[n\beta-it]^2+s^2 }\right)+K_0\left(\sqrt{[n\beta+it]^2+s^2}\right)\right]$ Let $\beta > 0$ and $t, s \in \mathbb{R}$. Furthermore, suppose that $-t^2 + s^2 > 0$. Define the following function:
$$
F( \beta, t, s )\ : = \ \sum_{n=1}^{\infty} \left[K_0\left(\sqrt{[n\beta-it]^2+s^2 }\right)+K_0\left(\sqrt{[n\beta+it]^2+s^2}\right)\right]
$$
where $K_0$ is the modified Bessel function of the second kind (McDonald function) of order $0$.
I know that $K_0(z) \sim - \log(z)$ for $z=0$ (this is where the function blows up). Also, I know that $K_0(z) \sim \frac{e^{-z}}{\sqrt{z}}$ for $z\to\infty$ where the function dies away to $0$.
I am interested in the following questions:


*

*Is there a way to evaluate this series? (I am guessing probably not. I have found nothing helpful in Gradshteyn+Ryzhik nor DLMF)

*How does this function look like in the limit $t \to 0$? (Asymptotically)

*How does this function look like in the limit $t \to \infty$?
I am more interested in the latter 2 questions. Because $F$ is a series I am not sure how to deal with this. 
 A: As $\Re\sqrt{\left( n\beta\pm it \right)^2+s^2}>0$, one may use the representation of $K_0$ in terms of the Hankel function:
\begin{equation}
 K_0\left(\sqrt{\left( n\beta\pm it \right)^2+s^2} \right)=\frac{i\pi}{2}H_0^{(1)}\left( i\sqrt{\left( n\beta\pm it \right)^2+s^2}  \right)
\end{equation} 
then, remarking that
\begin{equation}
  i\sqrt{\left( n\beta\pm it \right)^2+s^2}=\sqrt{\left( \mp t+in\beta \right)^2-s^2}
\end{equation} 
one have the integral representation
\begin{equation}
\left(\frac{z+\zeta}{z-\zeta}\right)^{\frac{1}{2}\nu}{H^{(1)}_{\nu}}\left((z^{%
2}-\zeta^{2})^{\frac{1}{2}}\right)=\frac{1}{\pi i}e^{-\frac{1}{2}\nu\pi i}\int%
_{-\infty}^{\infty}e^{iz\cosh u+i\zeta\sinh u-\nu u}\mathrm{d}u
\end{equation} 
when $\Im\left( z\pm\zeta \right)>0$. With $\nu=0,z_\pm=\mp t+in\beta,\zeta=s$, the condition is fulfilled. Thus, 
\begin{align}
 F(\beta,t,s)&=\sum_{n=1}^\infty \int_{-\infty}^\infty \cos\left( t\cosh u \right)e^{-n\beta\cosh u+is\sinh u}\,du\\
&=\int_{-\infty}^\infty \cos\left( t\cosh u \right)\frac{e^{-\beta\cosh u+is\sinh u}}{1-e^{-\beta\cosh u}}\,du\\
&=\int_{-\infty}^\infty \cos\left( t\cosh u \right)\cos\left( s \sinh u\right)\frac{e^{-\beta\cosh u}}{1-e^{-\beta\cosh u}}\,du\\
&=\Re\int_{-\infty}^\infty \cos\left( s \sinh u\right)\frac{e^{(it-\beta)\cosh u}}{1-e^{-\beta\cosh u}}\,du
\end{align} 
To obtain the asymptotic behavior for $t\to\infty$, using the Laplace method, one has
\begin{align}
 F(\beta,t,s)&\sim \Re\left[\frac{e^{it-\beta}}{1-e^{-\beta}}\int_{-\infty}^\infty e^{\left( it-\beta \right)u^2/2}du+O\left( t^{-3/2} \right)\right]\\
&\sim \frac{\sqrt{2\pi}}{e^{\beta}-1}\Re\left[\frac{e^{it}}{\sqrt{\beta-it}}\right]\\
&\sim \frac{\sqrt{\pi}}{e^{\beta}-1}\frac{\cos t}{\sqrt t}
\end{align}
Expansions with higher order terms (which involve the parameter $s$) can easily be obtained.  
The behavior of the function near $t=0$ using the obtained integral representation is not so simple as it require the calculation of difficult integrals. For example,
\begin{equation}
 F(\beta,0,s)=2\int_{0}^\infty\frac{ \cos\left( s \sinh u\right)}{e^{\beta\cosh u}-1}\,du
\end{equation} 
