calculus and analysis - Calculation of an integral with a Bessel function

How can I calculate below integral? Is it possible to calculate it in Mathematica? If yes, how? Am I doing something wrong?

Integrate[BesselJ[1, q x] x /(x^2 + a^2)^(5/2), {x, a, Infinity}]

Answer

It seems that for some values of $a$ greater for some value, the integral value is complex. I have done the following:

$\int_a^\infty \frac{x J_1(q x)}{(x^2+a^2)^{5/2}}dx=\int_0^\infty \frac{x J_1(q x)}{(x^2+a^2)^{5/2}}dx-\int_0^a \frac{x J_1(q x)}{(x^2+a^2)^{5/2}}dx$

The first term is a Hankel transform of order 1, the other can be done substituting the Bessel function by its series equivalent. Thus:

HankelTransform[1/(x^2 + a^2)^(5/2), x, q, 1]

$\frac{8 G_{3,1}^{1,2}\left(\frac{4}{q^2 a^2}| \begin{array}{c} -\frac{3}{2},-\frac{1}{2},\frac{1}{2} \\ 0 \\ \end{array} \right)}{3 \sqrt{\pi } \left(a^2\right)^{5/2} q^2}$

and

Assuming[a > 0 && a \[Element] Reals && m \[Element] Integers && m >= 0, 
Integrate[(q x)^(2 (m + 1)) (-1)^m/(2^(2 m + 1) Gamma[m + 2] m! (x^2 + a^2)^(5/2)), {x, 0, a}]]

with finaly:

$\sum _{m=0}^{\infty } \frac{i 2^{-2 (m+2)} q^{2 (m+1)} (-a)^{2 (m-1)} \left((4 m (4 m+5)-1) B_{-1}\left(m+\frac{3}{2},\frac{1}{2}\right)-2 (m+2) (4 m-1) B_{-1}\left(m+\frac{3}{2},\frac{3}{2}\right)\right)}{3 m! \Gamma (m+2)}$

Plotting the real part of their difference, with $m=80$ and $q=1$:

which agrees with the other approaches.

You can modify the value of $m$ to the required accuracy.