$k$-th moment of product of gaussian and sinc

Question

I would like to calculate the following integrals:

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

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

Thanks!

Answer 1

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.

Answer 2

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

can be solved using known integrals involving bessel function of the first kind.

Since $$\left(\frac{\sin(\pi a x)}{\pi ax}\right)^2=\frac{1}{2ax}J_{\frac{1}{2}}(\pi ax)J_{\frac{1}{2}}(\pi ax)$$

1. 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 $k\geq 0, \,a>0,\,b>0$.