Integrate $\int_{-\infty}^\infty\frac{e^{-ik\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx$ I'm trying to evaluate the integral below for my research related to sound radiation. Assume $a$ is a positive constant.
$$\int_{-\infty}^\infty\frac{e^{-ik\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx$$
First, I tried to separate the functon real and imaginary part.
$$\int_{-\infty}^\infty\frac{\cos({k\sqrt{x^2+a^2})}}{\sqrt{x^2+a^2}}dx-i\int_{-\infty}^\infty\frac{\sin({k\sqrt{x^2+a^2})}}{\sqrt{x^2+a^2}}dx$$
And I tried to contour integral using branch cut similar to the link. However I can't handle branch cuts on the imaginary line well.(Is the answer of the link only way to handle it?) How to handle that branch cuts simple as the case of real line. 
And.... How to evaluate the whole integral??

Thanks for several answers. I've thought the problem too hard..
However, I made a mistake to my question. I need to consider the case when $a=0$. I guess...
$$\int_{-\infty}^\infty\frac{e^{-ik\sqrt{x^2}}}{\sqrt{x^2}}dx=2\int_{0}^\infty\frac{e^{-ik\sqrt{x^2}}}{\sqrt{x^2}}dx=2\int_{0}^\infty\frac{e^{-ikx}}{x}dx$$
Is it valid?
 A: Your integrand is even, so rewrite as
$$
\int_{-\infty}^\infty\frac{e^{-ik\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx=2\int_{0}^\infty\frac{e^{-ik\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx\ ,
$$
then set $\sqrt{x^2+a^2}=t\Rightarrow x=\sqrt{t^2-a^2}$, yielding
$$
2\int_{a}^\infty\frac{e^{-ikt}}{t}\frac{2t dt}{2\sqrt{t^2-a^2}}=2\int_a^\infty\frac{e^{-i k t}dt}{\sqrt{t^2-a^2}}=2\int_1^\infty\frac{e^{-i k a t}dt}{\sqrt{t^2-1}}\ .$$
Mathematica can handle the resulting integral and gives the answer $2K_0(i a k)$, where $K_0$ is a Bessel function [checked numerically]. I'll try to see if I can find this integral on Gradshteyn.
EDIT The last integral is basically one of the integral representation of the Bessel function, see http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/07/01/01/0001/
A: THIS IS NOT AN ANSWER
putting the substitution $x = a\tan(\theta)$ greatly simplifies the integral. Thereafter, multiplying and dividing by $\tan(\theta)$ in the transformed integral, and again integrating by parts two times would give you the answer.
This may look a little lengthy but actually isn't.
