$$\int ^\infty _0 x J_0(k x)e^{-x^2/2}dx$$
The integral above corresponds to fourier transform in radial coordinates. The fourier transform of a 2D gaussian is still a 2D gaussian. So the integral above is $e^{-k^2/2}$
This can be shown using the series expansion of bessel functions somewhat easily, however I was wondering if there could be a way without getting into series. Or is there perhaps anything interesting to know about taking such integrals that might interest me?
Edit: More generally $$\int ^\infty _0 x J_n(k x)x^ne^{-x^2/2}dx=k^ne^{-k^2/2}$$
this is not a part of the question, just interesting
Ok, now that i have some time i will write down something:
First recall that
$$J_0(y)=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i y \sin(t)}dt$$
so that our integral can equivalently written after exchanging the $t$ and $x$ integration $$ 2\pi I(k)=\int_{-\pi}^{\pi} dt \int_0^{\infty} x e^{-x^2-ikx\sin(t)}dx=\\ \partial_k\int_{-\pi}^{\pi} dt\frac{1}{-i\sin(t)} \underbrace{\int_0^{\infty} e^{-x^2-ikx\sin(t)}dx}_{J(k,t)}\qquad(1)$$
The inner integral is easily calculated in terms of Error functions with imaginary argument (denoted $\text{Erfi}(x)$):
$$ J(k,t)=\frac{\sqrt{\pi }}{2}e^{-\frac{1}{4} k^2 \sin ^2(t)}-\frac{i\sqrt{\pi }}{2} \text{Erfi}\left(\frac{1}{2} k \sin (t)\right)e^{-\frac{1}{4} k^2 \sin ^2(t)} $$
Performing the $k$ derivative yields: $$ \frac{\partial_k J(k,t)}{-i\sin(t)} =\frac{1}{2}-\frac{i\sqrt{\pi } k}{4} \sin (t) e^{-\frac{1}{4} k^2 \sin ^2(t)} \left(1-i\text{Erfi}\left(\frac{1}{2} k \sin (t)\right)\right) \qquad(2) $$
now plugging in (2) in (1) and substituting $y=\sin(t)$ we get $$ 2\pi I(k)=\pi+\frac{i\sqrt{\pi } k}{4}\underbrace{\int_{-1}^{1} \frac{ y e^{-\frac{1}{4}k^2y ^2}}{\sqrt{1-y^2}}dy}_{ =\,0\,\, \text{by parity}}+\frac{\sqrt{\pi } k}{4}\underbrace{\int_{-1}^{1} \frac{ y e^{-\frac{1}{4}k^2y ^2}\text{Erfi}\left(\frac{1}{2}ky\right)}{\sqrt{1-y^2}}dy}_{C(k)} \quad (3) $$
To calculate $C(k)$ lets set $k=i \tilde{k} $ which should be causing now problems because our integrand is analytic in the complex plane with slit from $[-1,1]$ and therefore analytic continuation is allowed (we also used parity and a subsitution $y\rightarrow \frac{1}{2}\tilde{k}y$ in this step) .
$$ C(\tilde{k})=-\frac{4i}{\tilde{k}}\int_{0}^{\frac{1}{2}\tilde{k}}\frac{ y e^{y ^2}\text{Erf}\left(y\right)}{\sqrt{\frac{1}{^4}\tilde{k}^2-y^2}}dy $$
This integral is "well known" and can be found in formula 4.3.19 here and we get (after rotating back $\tilde{k}$) $$ C(k)=\frac{4}{k}\sqrt{\pi}\left(e^{-\frac{1}{4}k^2}-1\right) \quad (4) $$
Pluggling (4) into (3) we obtain
I would be really grateful if someone could provide a proof of the linked formula.