calculus - How to evaluate integral $int_0^{infty} e^{-x^2} frac{sin(a x)}{sin(b x)} dx$?

I came across the following integral:

$$\int_0^{\infty} e^{-x^2} \frac{\sin(a x)}{\sin(b x)} dx$$

while trying to calculate the inverse Laplace transform

$$L_p^{-1} \left[ \frac{\sinh(\alpha\sqrt{p})}{\sinh(\beta\sqrt{p})} \frac{e^{-\gamma\sqrt{p}}}{\sqrt{p}} \right], |\alpha|<\beta, \gamma>0$$

using the Bromwich integral approach. The contour I used is the following:

the above mentioned integral arises while doing integration over the segments $L_1^+,L_2^+,\cdots$ and $L_1^-,L_2^-,\cdots$.

I have searched for this integral in Prudnikov et. al., Integrals and Series, v.1, but found nothing. I have also tried to evaluate the integral using residue theorem, but could not quite decide which contour to use.

Any help is greatly appreciated!

P.S. The ILT can be calculated by noticing that
$$F[p] = \frac{\sinh (\sqrt{p} \alpha)}{\sinh (\sqrt{p} \beta)} \frac{e^{-\gamma\sqrt{p}}}{\sqrt{p}} = \sum_{n=0}^{\infty} \left( \frac{e^{-(-\alpha+\beta+\gamma+2n\beta)\sqrt{p}}}{\sqrt{p}} -\frac{e^{-(\alpha+\beta+\gamma+2n\beta)\sqrt{p}}}{\sqrt{p}} \right)$$
using
$$L_p^{-1} \left[ \frac{e^{-\alpha\sqrt{p}}}{\sqrt{p}} \right] = \frac{1}{\sqrt{\pi t}} e^{-\frac{\alpha^2}{4t}}$$
we get
$$\begin{align*} f(t) &= L_p^{-1}[F(p)] \\ &= \sum_{n=0}^{\infty} \left( \frac{ e^{-(-\alpha+\beta+\gamma+2n\beta)^2/4t} }{\sqrt{\pi t}} - \frac{ e^{-(-\alpha+\beta+\gamma+2n\beta)^2/4t} }{\sqrt{\pi t}} \right). \end{align*}$$
Here I am more interested in calculating the above ILT using the Bromwich integral approach.