I need compute the integral
$$\frac1{(2\pi)^{2} }\int d^2p \ e^{-l|p|}e^{i \vec{x} \cdot \vec{p}}$$
The problem does not specified the limits of integration
The result is
$$\frac{1}{2\pi} \frac{l}{(\sqrt{\vec{x}^2+l^2})^3}$$
I have tried using
$$\int_0^{2\pi} d\theta \int_0^\infty dp p e^{-lp} e^{ipx \cos (\theta)} $$
and my result using mathematica is: $$\frac{1}{2\pi} \int_0^\infty dp e^{-lp}p J_o(px)$$
But i dont know how to compute this integral to arrive to the results
\begin{align} &\frac1{(2\pi)^2}\int_{{\Bbb R}^2} e^{-l|p|}e^{i \vec{x} \cdot \vec{p}}\ d^2p\\ =& \ \frac1{(2\pi)^2}\int_0^{2\pi}\int_0^\infty e^{-lp +ixp\cos\theta}p dp \ d\theta \\ =& \ \frac1{(2\pi)^2}\int_0^{2\pi}\int_0^\infty p e^{-lp}\cos(xp\cos\theta) dp\ d\theta \\ =& - \frac1{(2\pi)^2} \frac d{dl}\int_0^{2\pi}\int_0^\infty e^{-lp}\cos(xp\cos\theta) dp\ d\theta \\ =&- \frac1{(2\pi)^2} \frac d{dl}\int_0^{2\pi} \frac{l}{l^2+x^2\cos^2\theta} \ d\theta\\ =& -\frac1{(2\pi)^2}\frac d{dl}\frac{2\pi}{(l^2+x^2)^{1/2} } = \frac l {2\pi(l^2+x^2)^{3/2}} \end{align}