What's the value of the integral given by : $$ I(a) = \int_0^\infty \dfrac{1}{x} e^{-a x} J_{3/2}(x)J_{3/2}(x) dx, $$ where $a$ is a positive real parameter.
I don't know if this could help, but following is the graph I obtain numerically.
Also, taking a look at Gradshteyn and Ryzhik (edition 8), no interesting formula seems to be proposed. Except for formula 6.626 (page 711), where it's mentioned that it applies only for $a > 1$. However, the numerical computation shows that the integral does converge for all $a$ as shown in the graph above.
Here is a direction which will give you a series solution:
Using formula 8.442 P. 960 of my edition of Gradshteyn and Ryzhik (Table of integrals, series and products, Academic Press, 1980) :
$$J_{3/2}(x)^2=\sum_{k=0}^{\infty}(-1)^k\left(\frac{x}{2}\right)^{2k+3}\frac{\Gamma(2k+4)}{k! \Gamma(k+4)\Gamma(k+\frac52)^2}$$
(take care to the last square). From there, integrate term by term to obtain a solution under the form of a series. Has this series a closed form ? Hopefully as an hypergeometric expression...