Compute $\int_0^\infty x^{n+1} J_n(kx) \exp(-a^2 x^2) dx$

Question

I need advice on how to solve the following integral:

$$\int_0^\infty x^{n+1} J_n(kx) \exp(-a^2 x^2) dx$$

Answer 1

First, make life simpler : let $x=\frac t k$ and $b=\frac{a^2}{k^2}$ which make $$I_n=\frac 1{k^{n+2}}\int_0^\infty t^{n+1}\,J_n(t)\,e^{-b\,t^2}\,dt$$ and, from the link given by @Gary $$I_n=\frac {\exp\left(-\frac 1{4b}\right) }{k^{n+2}(2b)^{n+1}}=\frac {k ^n}{2^{n+1}\, a^{2 (n+1)} } \exp\left(-\frac{k^2}{4 a^2}\right)$$

Answer 2

A possible approach to get it is to substitute directly the power series for $J_n$: \begin{align*} \int_0^\infty x^{n+1}J_n(kx)e^{-a^2 x^2}\,dx &=\int_0^\infty x^{n+1}e^{-a^2 x^2}\sum_{j=0}^\infty\frac{(-1)^j(kx/2)^{n+2j}}{j!\ \Gamma(n+j+1)}\,dx \\&=\sum_{j=0}^\infty\frac{(-1)^j(k/2)^{n+2j}}{j!\ \Gamma(n+j+1)}\int_0^\infty x^{2(n+j)+1}e^{-a^2 x^2}\,dx \\&=\sum_{j=0}^\infty\frac{(-1)^j(k/2)^{n+2j}}{j!\ \Gamma(n+j+1)}\frac{\Gamma(n+j+1)}{2a^{2(n+j+1)}} \\&=\frac{k^n}{(2a^2)^{n+1}}\sum_{j=0}^\infty\frac{(-1)^j}{j!}\left(\frac{k}{2a}\right)^{2j}=\frac{k^n}{(2a^2)^{n+1}}e^{-(k/2a)^2}. \end{align*}