I would like to establish the asymptotic behaviour of the following integral: $$ I(r) = \int_0^\infty dq \frac{q}{\sqrt{m^2 + q^2}} e^{-\sigma q^2 /2} \sin (q r) $$ for $r \to + \infty$, with $m\geq 0$ and $\sigma >0$ real parameters. I know that $\lim_{r \to \infty} I(r) =0$, but I would like to know how it goes there. For example, in the $m=0$ case the integral can be performed exactly (it's the sum of two Gaussian integrals with complex coefficients), and the result is: $$ I(r) \xrightarrow[m\to 0]{} \sqrt{\frac{\pi }{2 \sigma }} \text{erfi}\left(\frac{r}{\sqrt{2\sigma }}\right) e^{-\frac{r^2}{2 \sigma }} = \frac 1 r + \mathcal{O}\left( e^{-\frac{r^2}{2 \sigma }} \right) . $$ I'm not sure how to go about the $m>0$ case. The sine function obviously can be written as a sum of two complex exponentials, and if I want I can also repack everything into a single integral from $q = -\infty$ to $q= + \infty$ with a single exponential: $$ I(r) = \frac{1}{2i} \int_{-\infty}^\infty dq \frac{q}{\sqrt{m^2 + q^2}} e^{-\sigma q^2 /2} e^{iq r} , $$ but I can't use the stationary phase approximation as it is, because the function $q$ in the exponent has no stationary points. I can transform for example $q = x^3$ and then $x=0$ is a stationary point, and I can expand the square root in Maclaurin series $\frac{1}{\sqrt{m^2+x^6}}\sim \frac{1}{m}-\frac{x^6}{2 m^3} + \dots$, and I can integrate separately each term, but what I get is that each term results in a $e^{-\frac{r^2}{2 \sigma }}$ factor multiplied by increasing powers of $r$, which means my expansion is not giving me the dominant behaviour for $I(r)$.
Thinking about the problem in the complex plane, I see that the oscillating parts $e^{\pm i q r}$ have their steepest descent directions on the imaginary axis of $q$, but the integrand $ \frac{q}{\sqrt{m^2 + q^2}} e^{-\sigma q^2 /2} $ converges to zero as $|q| \to \infty$ only in the butterfly-shaped region $|Im(q)|<|Re(q)|$, which does not include the imaginary axis. Moreover the square root at the denominator gives two poles at $q =\pm im$, from which two branch cuts have to depart and go to infinity. I don't see a deformation of the path of the positive real axis that allows me to go along the steepest-descent path and exploit the fact that $r$ is large.
This is as far as I got in the analysis of this problem, any suggestion on the best way to go about this?
Thanks everyone and best regards.
Assume that $m > 0$ and $\sigma > 0$ are constant. The critical point of $-\sigma^2 q^2/2 + i r q$ is $q_0 = i r/\sigma$. If the non-exponential part didn't have a branch cut, the steepest descent contour would go through $q_0$ parallel to the real axis. But the contour for $I(r)$ has to be augmented with the segments from $q_0$ to the branch point $q_b = i m$ on both sides of the branch cut. The main contribution to the integral comes from a small neighborhood of $q_b$ and can be estimated by applying Laplace's method.
Taking $q = q_b + i \xi$ and expanding the exponential and non-exponential parts of the integrand around $\xi = 0$, we obtain $$I(r) = \frac 1 2 \,\operatorname{Im} \int_{-\infty}^\infty \frac q {\sqrt {q^2 + m^2}} e^{-\sigma q^2/2 + i r q} dq \sim e^{m (m \sigma/2 - r)} \int_0^\infty \sqrt {\frac m {2 \xi}} \,e^{(m \sigma - r) \xi} d\xi \sim \\ e^{m (m \sigma/2 - r)} \sqrt {\frac {\pi m} {2 r}}, \quad r \to \infty.$$