I want to find an estimate on the magnitude of the following function in terms of $x,y$ and $N$ when $N$ is large: $$ f(N) = \sum_{m\in \mathbb{Z}} H_{m-N}^{(1)}(x) J_m(y), \quad \quad (*) $$ where $H_{m}^{(1)}$ is the Hankel function of the first kind of order $m$ and $J_m$ is the Bessel funtion of order $m$ and $x>y>0$.
I am used to estimating Hankel and Bessel functions when $m$ is large and we can use \begin{align} |J_m(x)| & \sim \sqrt{\frac{1}{2 \pi m}} \bigg(\frac{ex}{2m}\bigg)^m, \\ |H_m^{(1)}(x)| & \sim \sqrt{\frac{2}{\pi m}} \bigg(\frac{ex}{2m}\bigg)^{-m}. \end{align}
In particular, this gives $|H_m^{(1)}(x)J_m(y)| \sim \frac{1}{m}\bigg(\frac{y}{x}\bigg)^m$ as $m \to \infty$.
However, it seems the above relations don't seem to be of any use for the more complicated expression $(*)$. Is there any way of getting an estimate on the function $f(N)$ as $N\to \infty$ or is this an impossible task?
From the parity formula \begin{equation} H^{(1)}_{-n}\left(z\right)=(-1)^{n}{H^{(1)}_{n}}\left(z\right)\quad ; \quad J_{-n}\left(z\right)=(-1)^{n}J_{n}\left(z\right) \end{equation} we can transform (*) as \begin{align} f(N) &= \sum_{m\in \mathbb{Z}} H_{m-N}^{(1)}(x) J_m(y)\\ &= (-1)^N\sum_{m\in \mathbb{Z}} H_{N-m}^{(1)}(-x) J_{-m}(-y)\\ &=(-1)^N\sum_{k\in \mathbb{Z}} H_{N+k}^{(1)}(-x) J_{k}(-y)\\ \end{align} Using the Neumann's addition theorem, \begin{equation} H^{(1)}_{\nu}\left(u- v\right)=\sum_{k=-\infty}^{\infty}H^{(1)}_{\nu+ k}\left(u\right)J_{k}\left(v\right) \end{equation} valid for $\left|u\right|>\left|v\right|$. Taking $u=-x,v=-y,\nu=N$ (and $x>y>0$) we obtain \begin{equation} f(N)=(-1)^NH^{(1)}_{N}\left(y-x\right) \end{equation} The asymptotic form for $N\to\infty$ is then \begin{equation} f(N)\sim (-1)^{N+1}\frac{i}{\sqrt{\pi}}2^{N+1/2}N^{N-1/2}e^{-N}\left( y-x \right)^{-N} \end{equation}