Does this integral have a closed form? $\int^{\infty}_{-\infty} \frac{e^{ikx} dx}{1-\mu e^{-x^2}}$ The original integral contains $\sin [n(x+a)] \sin [l(x+a)]$ but I think this form is simpler:
$$\int^{\infty}_{-\infty} \frac{e^{ikx} dx}{1-\mu e^{-x^2}}$$
$\mu <1$, so I can use the Taylor expansion to calculate the integral with very good accuracy.
$$\frac{1}{1-\mu e^{-x^2}}=1+\sum^{\infty}_{j=1} \mu^j e^{-jx^2}$$
And then of course I complete the square in the exponent and it becomes Poisson integral.
The series will be:
$$\sum^{\infty}_{j=1}\frac{ \mu^j}{\sqrt{j}} e^{-b^2/j}$$
($b$ depends on $k$, but in the original expression it is more complex, so its exact form doesn't matter, it's sufficient that it is real. )
If this series or the original integral have a closed form it will be more convenient. I wasn't able to find it in the literature.

We have:
$$\begin{eqnarray*}\int_{-\infty}^{+\infty}\frac{e^{ikx}}{1-\mu e^{-x^2}}\,dx = \int_{-\infty}^{+\infty}\frac{\cos(kx)}{1-\mu e^{-x^2}}\,dx &=& 2\sum_{r=1}^{+\infty}\mu^r\int_{0}^{+\infty}\cos(kx)e^{-rx^2}\,dx\\&=&2\sum_{r\geq 1}\mu^r \sqrt{\frac{\pi}{r}}\,e^{-\frac{k^2}{4r}}\\&=&2\sqrt{\pi}\sum_{r\geq 1}\frac{\mu^r}{\sqrt{r}}\,e^{-\frac{k^2}{4r}}\end{eqnarray*}$$
that can be re-written through the Poisson summation formula, but does not have a "nice" closed form.
