# double integral equal to modified Bessel function

I need to prove that, \begin{eqnarray} \frac{1}{2\pi}\int^{\infty}_{-\infty} \int^{\infty}_{-\infty} dk_x dk_z \left( \frac{\exp[i a k_x ]\exp[ib k_z]}{k^2_x + k_z ^2 + \frac{1}{l^2}}\right) = K_0\left(\frac{\sqrt{a^2+b^2}}{l} \right) \end{eqnarray}

$i$ is unit complex number and $K_0$ is Bessel function of second kind and one of the definition is

\begin{eqnarray} K_0(x) = \int^{\infty}_{0} \frac{\cos[x t]}{\sqrt{1+t^2}} dt \end{eqnarray}

I have checked the relation with Mathematica for number of values of $(a,b)$.

To proceed, I used result from wiki , \begin{eqnarray} \int^{\infty}_{-\infty} \frac{\exp[i a t]}{{p^2+t^2}} dt = \pi\frac{\exp[-pa]}{p} \end{eqnarray}

then, integrating w.r.t $k_x$ first \begin{eqnarray} \int^{\infty}_{-\infty} \left[\int^{\infty}_{-\infty} \left( \frac{\exp[i a k_x ]}{k^2_x + k_z ^2 + \frac{1}{l^2}}\right) dk_x \right] \exp[ib k_z] dk_z = \int^{\infty}_{-\infty} \frac{\exp[-pa]}{p} \exp[ib k_z] dk_z \end{eqnarray} where \begin{eqnarray} p^2 = k^2_z + \frac{1}{l^2} \end{eqnarray}

Now the integral turns ugly and I do not see how to obtain factor $\sqrt{a^2+b^2}$ together.

How can I proceed?

• Hint: switch to polar coordinates. – Jack D'Aurizio Feb 10 '15 at 18:15

Let we replace $k_x$ with $x$ and $k_z$ with $z$. Then we are looking for: $$\begin{eqnarray*} I &=& \frac{1}{2\pi}\iint_{\mathbb{R}^2}\frac{\exp\left(i(ax+bz)\right)}{x^2+z^2+\frac{1}{l^2}}\,dx\,dz\\ &=& \frac{1}{2\pi}\int_{0}^{+\infty}\int_{0}^{2\pi}\frac{\rho}{\rho^2+\frac{1}{l^2}}\exp\left(i\left(a\rho\cos\theta+b\rho\sin\theta\right)\right)\,d\theta\,d\rho\\&=&\int_{0}^{+\infty}\frac{\rho}{\rho^2+\frac{1}{l^2}}\cdot J_0\left(\rho\sqrt{a^2+b^2}\right)\,d\rho.\tag{1}\end{eqnarray*}$$ Now we can exploit the following identities related with the (inverse) Laplace transform: $$\mathcal{L}^{-1}\left(\frac{\rho}{\rho^2+\frac{1}{l^2}}\right)=\cos\left(\frac{s}{l}\right),$$ $$\mathcal{L}\left(J_0(K \rho)\right)=\frac{1}{\sqrt{K^2+s^2}}\tag{2}$$ to deduce: $$I = K_0\left(\frac{1}{l}\sqrt{a^2+b^2}\right)\tag{3}$$ as wanted.
• Thanks a lot Jack for the solution. Switching to polar coordinates I see where factor $\sqrt{a^2+b^2}$ comes from. I see that Laplace transforms you write are correct but how is Eq.(3) is derived from 2? – chatur Feb 11 '15 at 17:27
• @chatur: by using $(2)$ we have that the integral can be written as $$\int \frac{\cos(s/l)}{\sqrt{K^2+s^2}}\, ds$$ that is a value for $K_0$ by your second line. – Jack D'Aurizio Feb 11 '15 at 17:36