I tried to evaluate integral $$ I = \int \limits_{0}^{\infty}\frac{J_{\nu}(bx)x^{\nu + 1}dx}{(x^{2} + a^{2})^{\mu + 1}}, $$ but I failed. I can't represent Bessel function as a series, because corresponding summands don't converge. Then I tried to use representation $J_{v}(x) = \frac{1}{2 \pi i}\left(\frac{x}{2}\right)^{\nu}\int_{-\infty}^{0}t^{-\nu - 1}e^{t - \frac{x^{2}}{4t}}dt$, but this substitution leads to the integral of type $$ I = \int t^{-\nu - 1}e^{t}dt \int \limits_{0}^{\infty}e^{-a^{2}b^{2}\frac{y}{4t}}\frac{y^{\nu}dy}{(y + 1)^{\mu + 1}}, $$ which I can't evaluate (the equality $I = ...$ is understanding in sense of proportionality).
Can you help me?
Edit.
Hmmm. It seems that I done it.
Edit. Yes!! In is need to rewrite $\frac{1}{(x^{2} + a^{2})^{\mu + 1}} $ as $\frac{1}{\Gamma (\mu + 1)}\int \limits_{0}^{\infty}e^{-y(x^{2} + a^{2})}y^{\mu}dy$. Then all will be easy.
It is a special case of equation (6.60) in http://arxiv.org/abs/1207.5845 .