How to compute $\int_0^{\infty}x^2e^{-c x^2}J_0(x)dx, c>0$?
It is easy to compute this integral when we replace $x^2$ by $x$. Namely, $$\int_0^{\infty}xe^{-c x^2}J_0(x)dx = \frac{e^{-\frac{1}{4 c}}}{2c}$$ by using a series expansion for $J_0(x)$.
Today, I put that $x^2$ there, Mathematica still gives the result. It is not clear how this follows from using a similar series expansion.