Inverse 2D-Fourier Transform of $\frac{1}{k_y^2 + k_x^2 + ik_x\sigma}$ I need to invert the 2D-Fourier transform
$\hat{C}(\mathbf{k})  = \frac{1}{k_y^2 + k_x^2 + ik_x\sigma}$.
I have tried with the residue theorem to first deal with the $k_y$ part but that leaves me with something impossible for the $k_x$ part. Any chance this could be calculated analytically ? Thank you !
 A: Let's define a direct FT as $\displaystyle \hat f(k)=\int_{-\infty}^\infty f(x) e^{ikx}dx$,
then the inverse FT is defined as $\displaystyle f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty \hat f(k) e^{-ikx}dk$, and let's define
$$g(x, y,\sigma)=\frac{1}{(2\pi)^2}\int_{-\infty}^\infty\int_{-\infty}^\infty \frac{e^{-ik_xx-ik_yy}}{k_y^2+k_x^2+2ik_x\sigma}dk_xdk_y$$
$$=\frac{1}{(2\pi)^2}\int_{-\infty}^\infty \int_{-\infty}^\infty\frac{e^{-ik_xx-ik_yy}}{k_y^2+(k_x+i\sigma)^2+\sigma^2}dk_xdk_y=\frac{1}{(2\pi)^2}\int_{-\infty}^\infty \int_{-\infty+i\sigma}^{\infty+i\sigma}\frac{e^{-ik_xx-ik_yy}e^{-x\sigma}}{k_y^2+k_x^2+\sigma^2}dk_xdk_y$$
Considering the integral in the complex plane (over $k_x$) along the closed rectangular contour $-R\,\to\,R\,\to\,R+i\sigma\,\to\,-R+i\sigma\,\to\,R\,$ we see, that the function $\displaystyle \frac{e^{-ik_xx-ik_yy}e^{-x\sigma}}{k_y^2+k_x^2+\sigma^2}$ does not have poles inside this contour (the poles are $k_x=\pm\sqrt{k_y^2+\sigma^2}$). We can also show that the integrals along $R\,\to\,R+i\sigma$ and $-R+i\sigma\,\to\,-R$ tend to zero as $R\to\infty$.
Therefore,
$$\oint\frac{e^{-ik_xx-ik_yy}e^{-x\sigma}}{k_y^2+k_x^2+\sigma^2}dk_x=0$$
and
$$g(x, y,\sigma)=\frac{e^{-x\sigma}}{(2\pi)^2}\int_{-\infty}^\infty \int_{-\infty}^{\infty}\frac{e^{-ik_xx-ik_yy}}{k_y^2+k_x^2+\sigma^2}dk_xdk_y\tag{1}$$
From the equation$\,\,\displaystyle \Big(\frac{\partial^2}{\partial x^2}+\displaystyle \frac{\partial^2}{\partial y^2}+\displaystyle \frac{\partial^2}{\partial z^2}\Big)\frac{1}{\sqrt{x^2+y^2+z^2}}=-4\pi\delta(x)\delta(y)\delta(z)$
follows
$$\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{ik_xx+ik_yy+ik_zz}}{\sqrt{x^2+y^2+z^2}}dxdydz=\frac{4\pi}{k_x^2+k_y^2+k_z^2}\tag{2}$$
From (1) and (2) we get
$$g(x, y,\sigma)=\frac{e^{-x\sigma}}{4\pi}\int_{-\infty}^\infty\frac{e^{i\sigma z}}{\sqrt{x^2+y^2+z^2}}dz$$
Using the evaluation of the integral (here)
$$g(x, y,\sigma)=\frac{e^{-x\sigma}}{2\pi}K_0(|\sigma|\sqrt{x^2+y^2})$$
To get the initial integral we change $\sigma\to\frac{\sigma}{2}$.
