I am trying to solve the following integral:
$$I(a,R,d)=\int_0^{+\infty}\,k^{-1}\,J_1(kR)\,J_1(ka)\,\exp(-d^2k^2)\,dk$$
where $J_1(x)$ is the Bessel function of the first kind and $R,a,d$ have real positive values.
Playing with some numbers it's easy to check that the integral is convergent (see here).
I tried to apply an integral from G&R's book (page 707, 6.633/1):
$$\int_0^{+\infty}\,x^{\lambda+1}\,J_{\mu}(\beta x)\,J_{\nu}(\gamma x)\,\exp(-\alpha x^2)\,dx=\frac{\beta^{\mu}\gamma^{\nu}\alpha^{-\frac{\nu+\mu+\lambda+2}{2}}}{2^{\nu+\mu+1}\Gamma(\nu+1)}\sum_{m=0}^{+\infty}\frac{\Gamma(m+1+0.5(\nu+\mu+\lambda))}{m!\Gamma(m+\mu+1)}\left(-\frac{\beta^2}{4\alpha}\right)^m\,F\left(-m,-\mu-m;\nu+1;\frac{\gamma^2}{\beta^2}\right)$$
valid for $Re(\nu+\mu+\lambda)>-2$.
In my case $\nu=\mu=1$ and $\lambda=-2$ (that is $Re(\nu+\mu+\lambda)=0>-2$), therefore I have:
$$I(a,R,d)=\frac{aR}{8d^2}\sum_{m=0}^{+\infty}\frac{1}{(m+1)!}\left(-\frac{R^2}{4d^2}\right)^m\,F\left(-m,-1-m;2;\frac{a^2}{R^2}\right)$$
Does anyone know how to simplify further the solution???
A simmetry issue ($\nu=\mu$) allows me to state that (does it really help?):
$$\frac{aR}{8d^2}\left(-\frac{R^2}{4d^2}\right)^m\,F\left(-m,-1-m;2;\frac{a^2}{R^2}\right)=\frac{aR}{8d^2}\left(-\frac{a^2}{4d^2}\right)^m\,F\left(-m,-1-m;2;\frac{R^2}{a^2}\right)$$
Thanks!
A closed form for the integral involves a two variables hypergeometric series (Attachment). This is a special function of the Appell hypergeometric kind. I doubt that a simpler special function be convenient. Unfortunately, this is not a breakthrough compared to your own approach.