Devising an analytical approach to solving two infinite (improper) integrals of type $\int_0^\infty f(u) J_\mu (au) J_\nu(bu) \,\mathrm{d}u$ While I was elaborating on a fluid mechanical problem, I came across the two following functions defined in terms of non-trivial infinite (improper) integrals over the wavenumber $u$ in the interval $(0,\infty)$
$$
\psi_\alpha (a,b) = \int_0^\infty \frac{u^\frac{3}{2}}{\left( u^2+\alpha^2 \right)^{\frac{1}{2}}} \, J_0 (au) J_{\frac{3}{2}}(bu) \,\mathrm{d}u
$$
and
$$
\phi_\alpha (a,b) = \int_0^\infty \left( \frac{u^2+\alpha^2}{u} \right)^{\frac{1}{2}} J_1 (au) J_{\frac{5}{2}}(bu) \,\mathrm{d}u
$$
wherein $\alpha > 0$ is a small parameter, and $a>0$ and $b>0$ are the variables.
Those two improper integrals arose while solving dual integral equations using the well-established analytical approaches devised by $\,$ S n e d d o n $\,$ and $\,$ C o p s o n.
It can easily be demonstrated that those two integrals are convergent.
In particular, for $\alpha = 0$, the corresponding expression can readily be determined analytically.
They are found to depend on whether $a<b$ or $a>b$.
In the general case of interest, however, I am unable to figure out how to proceed. I have tried to use the series or integral representations of the Bessel functions and try to make an analytical progress but unfortunately this did not seem to work. 
Any help is highly appreciated.
Thank you very much!

N O T E : Using the change of variables $v=u/\alpha$, $A=\alpha a$, and $B=\alpha b$, the integrands can me made independent of the parameter $\alpha$.


Poisson’s and Related Integrals (10.9.3 in DLMF)
$$
J_\nu(z) = \frac{2 \left( \frac{z}{2} \right)^\nu}{\pi^\frac{1}{2} \Gamma \left( \nu + \frac{1}{2} \right)} 
\int_0^1 \left( 1-t^2 \right)^{\nu-\frac{1}{2}} \cos (zt) \, \mathrm{d}t \, .
$$

 A: I tried sketching on the start of a way forward with the Mellin transform for the first integral, i.e.
$$
\int_0^{\infty} \frac{t^{3/2}}{(t^2+\alpha^2)^{1/2}}J_0(at)J_{3/2}(bt)dt
$$
Note that this is far from being an actual solution at the moment. It was rather too long to fit into the comments. I'm not sure all the terms I got are correct, I haven't double checked them, but what still remains is the inverse Mellin transform which really doesn't look like too much fun, even though I think it is possible to compute. Maybe someone can take up the calculations to compute the inverse Mellin transform?
I started with trying to write this as an integral on the form of
$$
\int_0^{\infty}h_1\left(\frac{x}{t}\right)h_2(t)\frac{dt}{t}
$$
as the Mellin transform of this is $\tilde{h}_1(s)\tilde{h}_2(s)$ where the two are the Mellin transforms of $h_1(t)$ and $h_2(t)$ respectively.
With $h_1(t) = J_{\nu}\left(\frac{2}{\sqrt{t}}\right)$ and $h(t) = f(t)J_{\mu}(at), f(t)=\frac{t^{3/2}}{(t^2+\alpha^2)^{1/2}}$. the Mellin transform of $h_1(t)$ is
$$
\mathcal{M}[h_1(t);s] =\frac{\Gamma(\nu/2-s)}{\Gamma(1+\nu/2-s/2)}
$$
so to get $J_\nu(bt) = h_1(2bt^{-2})$, one has
$$
\mathcal{M}[J_\nu(bt);s] = \mathcal{M}[h_1(2bt^{-2});s] = 2^{-s/2-1}b^{-s/2}\frac{\Gamma(\nu/2-s/2)}{\Gamma(s/2+\nu/2+1)}
$$
with $\nu=3/2$, this simplifies to $\mathcal{M}[J_\nu(bt);s] = 2^{-s/2-1}b^{-s/2}\frac{\Gamma(3/4-s/2)}{\Gamma(s/2+5/2)}$ with the fundamental strip $-5<\Re(s)<3/2$.
For $h_2(t)$, the Mellin transform is in turn
$$
\tilde{h}_2(s) = \int_0^{\infty} t^{s-1}f(t)J_0(at)dt = \int_0^{\infty} \frac{t^{s+1/2}}{(t^2+\alpha^2)^{1/2}} J_0(at)dt
$$
So this is also on the form of a Mellin transform. But this integral is luckily given in 6.567.8 (p. 679) in Table of integrals, series and products, Gradshteyn, Ryzhik, 7th ed. in the form of
$$
\int_0^{\infty} \frac{x^{\rho-1}J_{\nu}(ax)}{(x^2+k^2)^{\mu+1}}dx
$$ which is the sum of two hypergeometric functions of ${}_1F_2$. For the parameters in this case, one has that
$$
\tilde{h}_2(s) = \int_0^{\infty} \frac{t^{s+1/2}}{(t^2+\alpha^2)^{1/2}} J_0(at)dt = \frac{\alpha^{s-3/2}\Gamma(s/2+3/4)\Gamma(-s/2-1/4)}{2\sqrt{\pi}}{}_1F_2\left(\frac{s}{2}+\frac{3}{4};\frac{s}{2}+\frac{5}{4},1;\left(\frac{a\alpha}{2}\right)^2 \right) + \frac{a^{-s-1/2}\Gamma(s/2+1/4)}{2^{-s+1/2}\Gamma(3/4-s/2)}{}_1F_2\left(\frac{1}{2};\frac{3}{4}-\frac{s}{2},\frac{5}{4}-\frac{s}{2};\left(\frac{a\alpha}{2} \right)^2 \right)
$$
with $-3/2<\Re(s)<-1/2$ if I've managed to put in all terms correctly. Now, as  $\tilde{h}_1(s)\tilde{h}_2(s)$ is known, it is "only" a matter of inverting this. And well, if the above is correct, one should then get whatever the original integral computes to. It looks like one would get infinite sums of the ${}_1F_2$ evaluated at each residual coming from the Gamma functions.
