I am reading a paper which has a number of two-dimensional Fourier transforms in the appendix. For example,
$$F\left(\frac{x^2}{r^3}\right)=\frac{k_2^2-k_1^2 z k}{k^3}e^{-kz},$$
where $r^2 = x^2 + y^2 + z^2$, $k^2 = k_1^2 + k_2^2$ and the 2D Fourier transform is defined as
$$\hat{f}(k_1, k_2, z)=F(f)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y,z) e^{ik_1x +ik_2 y} \: dx \: dy.$$
I suspect that these were obtained by treating the Fourier transform as a contour integral in the upper half-plane and was wondering if someone could carry out the integral and show how to do this to get the above result as an example of the method.
Edit: I see how an integral from $-\infty$ to $\infty$ of a function of $x$ with respect to $x$ can be written as a contour integral along the real axis and a semicircle in the upper half plane if we assume $f$ to be analytic in the upper half-plane except for a finite number of poles.
Here we have a 2D Fourier transform which is defined to be an integral of a function of $3$ variables where the integral is with respect to the first $2$ variables, so how do you write this as a contour integral in the upper half-plane?
Best Answer
At first your functions are not meromorphic, you can't solve them directly with contour integrals.
The solution is formula 504 on wikipedia
With $k=(k_x^2+k_y^2)^{1/2}$ $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-z (x^2+y^2)^{1/2}} e^{i(k_x x+k_yy)}dxdy=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-z (x^2+y^2)^{1/2}} e^{ik y}dxdy$$ $$=\int_0^{2\pi}\int_0^\infty e^{-z r} e^{i k r\sin t}rdrdt=\int_0^{2\pi}\frac1{(z-ik\sin t)^2} dt = \int_{|s|=1} \frac1{(z-ik \frac{s-s^{-1}}{2i})^2}\frac{ds}{is}$$ $$= 2i\pi (Res( \frac1{(sz-k \frac{s^2-1}{2})^2}\frac{1}{is},s=0)+Res( \frac1{(sz-k \frac{s^2-1}{2})^2}\frac{1}{is},s=\frac{-z+ (z^2+k^2)^{1/2}}{-k})+Res( \frac1{(sz-k \frac{s^2-1}{2})^2}\frac{1}{is},s=\frac{-z- (z^2+k^2)^{1/2}}{-k}))$$ $$= \ldots$$