I would like to know the leading order behaviour of
$$\sum_{m=-\infty}^{\infty}\frac{e^{i\left(\alpha m+\beta\sqrt{R^{2}+(x-mp)^2}\right)}}{\sqrt{R^{2}+(x-mp)^2}}$$
as $R/p\to\infty$ (first term in the asymptotic expansion is enough). It may be assumed that all variables are real-valued. Thanks in advance.
I've managed to work out the answer using the stationary phase method and I've checked it numerically in Mathematica. For interest, it is $$ \sum_{m=-\infty}^{\infty}\frac{e^{i\left(\alpha m+k\sqrt{R^{2}+(x-mp)^{2}}\right)}}{\sqrt{R^{2}+(x-mp)^{2}}}\sim\begin{cases}\frac{2^{1/2}\pi^{1/2}e^{i\pi/4}}{p^{1/2}R^{1/2}}\sum_{m=\lceil\frac{-kp-\alpha}{2\pi}\rceil}^{\lfloor\frac{kp-\alpha}{2\pi}\rfloor}\frac{e^{i\left(\frac{(\alpha+2\pi m)x+R\sqrt{k^{2}p^{2}-(\alpha+2\pi m)^{2}}}{p}\right)}}{(k^{2}p^{2}-(\alpha+2\pi m)^{2})^{1/4}},\space-kp<\alpha<kp\\\frac{2^{1/2}\pi^{1/2}e^{\frac{i\alpha x-R\sqrt{\alpha^{2}-k^{2}p^{2}}}{p}}}{p^{1/2}R^{1/2}(\alpha^{2}-k^{2}p^{2})^{1/4}},\space|\alpha|>kp\end{cases} $$ There is at least one propagating mode in the first case whereas there are only evanescent modes in the second case (where only the slowest decaying evanescent mode need be included).