Can somebody evaluate the following Gaussian integral?
$$ I(t,\sigma) := \int_{-\infty}^\infty \frac{dx e^{-x^2/(2\sigma^2)}}{\sqrt{2\pi \sigma^2}} \frac{\sin{\left(2 t\sqrt{1+x^2} \right )}}{\sqrt{1+x^2}} \tag{1} $$
Where of course $\sigma>0$ (and $t\in\mathbb{R}$). Expanding the $\sin$ I could not resum the series. However I have some hope (perhaps misplaced) that the integral can be expressed in terms of elementary functions.
In fact an analogous integral (that arises in the same problem) turns out to be surprisingly simple:
\begin{align} G(t,\sigma) &:= \int_{-\infty}^\infty \frac{dx e^{-x^2/(2\sigma^2)}}{\sqrt{2\pi \sigma^2}} \tag{2} \frac{1+x^2\cos{\left(2 t\sqrt{1+x^2} \right )}}{1+x^2}\\ &\simeq \exp\left( -2 \sigma^2 \sin(t )^2 \right) \end{align}
Added
My memory failed me in a more complicated way. It turns out that the above equation is not an exact evaluation of the integral $G$ but instead an excellent approximation (based on some physical theory).
To my excuse here is a plot of $G(t,\sigma=0.2)$ (dots) versus the approximation (continuous line)
At this point I have little hope that $I$ (or $G$) can be still evaluated analytically but I would love to be proved wrong.
Thanks again!
I managed to get this into a series, but not a closed form. I got that $$I = \sqrt{t\pi}\sum_{k=0}^{\infty}\left(-\frac{\sigma^2t}{2}\right)^k \binom{2k}{k}J_{1/2 + k} (2t)$$
The question is to solve
$$\int_{-\infty}^\infty \frac{e^{-x^2/(2\sigma^2)}}{\sqrt{2\pi \sigma^2}} \frac{\sin{\left(2 t\sqrt{1+x^2} \right )}}{\sqrt{1+x^2}} dx$$
The constant $\frac{1}{\sqrt{2\pi\sigma^2}}$ can be taken out of the integral to get
$$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} \frac{\sin{\left(2 t\sqrt{1+x^2} \right )}}{\sqrt{1+x^2}} dx$$
Here the $\sin$ can be expanded into a sum to get
$$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} \frac{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(2 t\sqrt{1+x^2} \right )^{2n+1}}{\sqrt{1+x^2}} dx$$
Swapping the sum and the integral, I get
$$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} \frac{\frac{(-1)^n}{(2n+1)!}\left(2 t\sqrt{1+x^2} \right )^{2n+1}}{\sqrt{1+x^2}} dx$$
The constants with respect to $x$ can be taken out of the integral
$$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} (1+x^2)^n dx$$
Here $(1+x^2)^n$ can be expanded to get
$$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} \sum_{k=0}^n \binom{n}{k} x^{2k} dx$$
The sum inside the integral can be taken out to get $$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\sum_{k=0}^n \binom{n}{k}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} x^{2k} dx$$
The inner integral can be calculated in closed form, so this simplifies to
$$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\sum_{k=0}^n \binom{n}{k}\sigma^{1 + 2k} \frac{(2k)!}{2^kk!}\sqrt{2\pi}$$
This reduces to $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\sum_{k=0}^n \binom{n}{k}\sigma^{2k} \frac{(2k)!}{2^kk!} \tag 1$$
Changing the order of summation from $n, k$ to $k, n$, I get $$\sum_{k=0}^{\infty}\sigma^{2k} \frac{(2k)!}{2^kk!}\sum_{n=k}^\infty \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1} \binom{n}{k}$$
Mathematica tells me that the inner sum can be written as $\frac{(-1)^k \sqrt{\pi} t^{1/2 + k} J_{1/2 + k} (2t)}{k!}$ where $J$ is the Bessel function of the first kind. This leads to $$\sqrt{t\pi}\sum_{k=0}^{\infty}\left(-\frac{\sigma^2t}{2}\right)^k \binom{2k}{k}J_{1/2 + k} (2t)$$