$k$-th moment of product of gaussian and sinc I would like to calculate the following integrals:


*

*$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx$$

*$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x\pm\pi)}{\pi ax\pm\pi}\right)^2\quad \exp(-bx^2) \,dx$$
Thanks!
 A: First one:


*

*The only non-zero moments correspond to even $k=2m$. In this case we have
$$I_k=\int_{-\infty}^{\infty}x^k\left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx=\frac{(-1)^{m-1}}{\pi^2a^2}\frac{\partial^{m-1}}{\partial b^{m-1}}\int_{-\infty}^{\infty}\sin^2\pi a x\, e^{-bx^2}dx$$

*But the last integral can be written as 
\begin{align}
\int_{-\infty}^{\infty}\sin^2\pi a x\, e^{-bx^2}dx=\frac14\int_{-\infty}^{\infty}\left(2-e^{2\pi i a x}-e^{-2\pi i a x}\right)e^{-bx^2}dx=\\
=\frac12\sqrt{\frac{\pi}{b}}\left(1-e^{-\pi^2a^2/b}\right),
\end{align}
where at the last step we have used the gaussian integral $\displaystyle \int_{-\infty}^{\infty}e^{-\beta x^2+2\alpha x}dx=\sqrt{\frac{\pi}{\beta}}\,e^{\alpha^2/\beta}$.


Therefore one finds
$$I_{2m}=\frac{(-1)^{m-1}}{2\pi^2a^2}\frac{\partial^{m-1}}{\partial b^{m-1}}\left[\sqrt{\frac{\pi}{b}}\left(1-e^{-\pi^2a^2/b}\right)\right].$$
Concerning the integrals of the 2nd type, consider the change of variables $y=x\pm a^{-1}$ and try adapt the above, it's not difficult.
A: *

*$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx$$


can be solved using known integrals involving bessel function of the first kind.
Since
$$\left(\frac{\sin(\pi a x)}{\pi ax}\right)^2=\frac{1}{2ax}J_{\frac{1}{2}}(\pi ax)J_{\frac{1}{2}}(\pi ax)$$


*

*can be written as


$$\int_{-\infty}^{+\infty} \quad\frac{1}{2a} x^{k-1}\,J_{\frac{1}{2}}(\pi ax)J_{\frac{1}{2}}(\pi ax)\, \exp(-bx^2)\,dx$$
That is a known integral from Volume II of "Higher Transcendental Functions". It holds for $k\geq 0, \,a>0,\,b>0$.
