Expectation of Sinc of Standardnormal Distribution Does anyone happen to know how to compute the expectation of the Sinc of a standard normal random variable, i.e.
$\mathbb{E} \Big[ \frac{\sin X}{X} \Big]$ where $X \sim \mathcal{N}(0,1)$?
Thanks!
EDIT: I suppose I could have elaborated what I have tried, so here goes:

*

*I tried to compute $ \int_\mathbb{R} \frac{\sin x}{x} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} \ dx$ but this didn't get me anywhere. I guess I didn't have a good approach as to how to calculate this integral.

*Also, I tried using $\mathbb{E} \Big[ \frac{\sin X}{X} \Big] = \mathbb{E} \Big[ \sum_{n = 0}^{\infty} \frac{(-1)^n X^{2n}}{(2n+1)!} \Big]$, applying dominated convergence (where I didn't check yet whether I can use it here because I wanted to see first where it leads) to pull in the expectation and then use the formula for the moments of the normal distribution. However, this didn't yield a (good) result either.

*Further, I tried first calculating the square of the integral in 1 first which is equal to $ \frac{1}{2 \pi} \int_\mathbb{R} \int_\mathbb{R} \frac{\sin x}{x} \frac{\sin y}{y}  e^{-(y^2 + x^2)/2}    \ dx \ dy$ and then try using polar coordinates but that was quite complicated and didn't help.

EDIT 2:
I tried another approach using the sine representation $ \sin x = \frac{e^{ix} - e^{-ix}}{2i}$ to compute the  integral in 1  which I was able to simplify as $ \frac{1}{\sqrt{2\pi} e^{1/2}} \int_{\mathbb{R}} \frac{1}{2ix} \Big( e^{-(x-i)^2/2 } - e^{-(x+i)^2/2 } \Big)$. Unfortunately, I am not sure how to compute this further.

Using the series for $\sin(x)$,
\begin{align}
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\sin(x)}{x}e^{-\frac{x^2}{2}}\,dx&=\sum_{k\ge 0}\frac{(-1)^k}{(2k+1)!}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^{2k}e^{-\frac{x^2}{2}}\,dx \\
&=\sum_{k\ge 0}\frac{(-1)^k (2k-1)!!}{(2k+1)!}=\sqrt{\frac{\pi}{2}}\left(\operatorname{erf}\!\left(\frac{\sqrt{2}}{2}\right)\right),
\end{align}
where the second equality follows from the fact that for $X\sim N(0,1)$, $\mathsf{E}X^{2k}=(2k-1)!!$, and the third equality uses the error function's Maclaurin series.
