Asymptotic behavior of $\int_0^\infty \rho\, d\rho \int_0^{2\pi} d \phi\ \frac{e^{i\mathbf \rho r \cos\phi}}{\rho + a}$ I would like to derive the asymptotic behavior of the following integral
$$
V_a(r)=\frac{1}{(2\pi)^2}\int_0^\infty \rho\, d\rho \int_0^{2\pi} d \phi\ \frac{e^{i\mathbf \rho r \cos\phi}}{\rho + a}
$$
for large $r>0$ and $a$ a fixed positive parameter. (This expression gives a screened Coulomb potential of an impurity in a two-dimensional electron gas). Recognizing the integral representation of the Bessel $J_0$ function, 
$$
\frac{1}{2\pi}\int_0^{2\pi} e^{i\rho r \cos\phi} d\phi = J_0(\rho r)\,,
$$
we have 
$$
V_a(r) = \frac{1}{2\pi r} \int_0^\infty \frac{x J_0(x)}{x+ar}dx
$$
and Mathematica gives
$$
V_a(r)=\frac{1}{2\pi r} + \frac{a}{4}\left(Y_0(ar)- \mathbf H_0(ar) \right) \underset{ar\to\infty}{\sim} \frac{1}{2\pi a^2 r^3}\,.
$$
I was wondering whether there is a strategy to obtain this asymptotic (or at least the $r^{-3}$ scaling) with some judicious approximation. I tried employing a stationary phase approximation, namely
$$
\int_0^{2\pi} e^{i\rho r \cos\phi} d\phi \approx \sqrt{\frac{2\pi}{i\rho r}}e^{i\rho r}\,,\text{ for large }\rho r
$$
but this introduces a non-integer scaling in $r$ so my guess is that it is not a correct strategy; furthermore, the parameter of this expansion is $\rho r$ and this cannot be always large since $\rho$ goes all the way down to $0$.
On account of this, I tried introducing a small cutoff $\eta>0$ in the the $\rho$ integration and then to estimate the error committed in doing so, but also this strategy keeps introducing fractional powers of $r$, together with other integrals I am not able to calculate.
 A: $$
V_a(r) = \frac{1}{2\pi r} \int_0^\infty \frac{x J_0(x)}{x+ar}dx
\qquad \text{OK}$$
$h(q)=2\pi\int_0^\infty \frac{x J_0(2\pi q x)}{x+c}dx\quad$ is the Hankel transform of  $\frac{1}{x+c}\quad$ with $c=ar$ and $q=\frac{1}{2\pi}$
$Y_0$ is the Bessel function of second kind of order 0, also refried as Neumann function. 
$\mathbf H_0$ is the Struve function of order 0.
$$V_a(r)=\frac{1}{2\pi r} + \frac{a}{4}\left(Y_0(ar)- \mathbf H_0(ar) \right)\qquad \text{OK}$$
An asymptotic expansion at $x\to\infty$ exists for the difference between Struve and Neumann functions of common argument and order (From Eq.[57:6:8] , p.566, in "An Atlas of Functions", J.Spanier, K.B.Oldham, Edit.1987, Hemisphere Pub. Corp., Springer-Verlag).
$$\mathbf H_0(x)-Y_0(x)\sim \frac{2}{\pi}\left( \frac{1}{x} -\frac{1}{x^3}+\frac{9}{x^5}+O\left(\frac{1}{x^6} \right) \right)$$
With $x=ar$ and $r\to\infty$ :
$$V_a(r)\sim\frac{1}{2\pi r} - \frac{a}{4} \frac{2}{\pi}\left( \frac{1}{ar} -\frac{1}{(ar)^3}+O\left(\frac{1}{r^4} \right) \right)$$
After simplification :
$$V_a(r)\sim  \frac{1}{2\pi a^2r^3}+O\left(\frac{1}{r^4} \right) $$
