I read in a book chapter by U. Frisch (Wave Propagation in Random Media) which states the 3D free space Green's function in the spatial domain as:
$$G(r,r')=\frac{\exp\big(ik(r-r')\big)}{-4\pi(r-r')}$$
And in the frequency domain (after Fourier Transform) as:
$$G(k)=({k_0}^2-k^2)^{-1}$$
I am trying to do the same operation with the 2D Green's Function which contains a Hankel operator to obtain a formulation in the frequency domain:
$$G_{2\mathrm D}(r)=\frac{i}{4}H_{0}^{(1)}(k_0r)$$
Something I have tried is to obtain a simplified formulation of the Green's function from https://dlmf.nist.gov/10.2 (equation 10.2.5):
$$H^{(1)}_{\nu}(z)\sim\sqrt{2/(\pi z)}~\mathrm{e}^{\mathrm i\left(z-\frac{1}{2}\nu \pi-\frac{1}{4}\pi\right)}$$
Then, I tried doing the Fourier transform operation for this simplified Hankel function, but to no avail as integration by parts does not converge to a solution due to the sqrt(1/z) term.
Does anyone have any suggestions on how I can Fourier Transform the 2D Green's function? Thank you.
EDIT: Eldar has suggested the use of a computational approach to obtain the result. We observe a discrepancy when using MATLAB and Wolfram to do the analysis and that is most likely due to the convention used between the 2 softwares in the Fourier Transform operation.
Best Answer
When I try to make a Fourir Transform (with respect to $z$) of your simplified function:
$$H(v,z)=\sqrt{\frac{2}{\pi z}} e^{i \left(-\frac{\pi v}{2}+z-\frac{\pi }{4}\right)}$$
then by a computational approach using
H[v_,z_]:=Sqrt[2/(Pi*z)]*E^(I(z-v*Pi/2-Pi/4)); FourierTransform[H[v,z],z,w], I obtain:$$\frac{(1+i) e^{i \left(-\frac{\pi v}{2}-\frac{\pi }{4}\right)} (\text{sgn}(\omega +1)+1)}{\sqrt{2 \pi } \sqrt{\left| \omega +1\right| }}$$
The transformation (with frequency variable $\omega$) includes the following sum of both, the cosine transform for the even and the sine transform for the odd part:
$$ \large\begin{array}{l} F_x\left(\sqrt{\frac{2}{\pi x}}e^{i\left(-\frac{\pi v}{2}+x-\frac{\pi}{4}\right)}\right)(\omega)=\\ F_x^c\left(-\frac{(-1)^{\frac{3}{4}} \left(\sqrt{-\frac{1}{x}}+e^{2ix}\sqrt{\frac{1}{x}}\right)}{\sqrt{2 \pi}e^{\frac{1}{2}i(\pi v+2x)}}\right)(\omega)+i\cdot F_x^s\left(\frac{(-1)^{\frac{3}{4}}\left(\sqrt{-\frac{1}{x}}-e^{2ix}\sqrt{\frac{1}{x}}\right)}{\sqrt{2\pi}e^{\frac{1}{2}i(\pi v+2x)}}\right)(\omega) \end{array} $$
The Fourier cosine transform for the even part is:
$$ \large\begin{array}{l} \sqrt\frac{2}{\pi}\int_{0}^{\infty}-\frac{(-1)^{\frac{3}{4}}e^{-\frac{1}{2}i(\pi v+2x)}\left(\sqrt{-\frac{1}{x}}+e^{2ix}\sqrt{\frac{1}{x}}\right)\cos(\omega x)}{\sqrt{2\pi}}\,dx=\\ \frac{\left(-\frac{1}{2}-\frac{i}{2}\right)(-1)^{\frac{3}{4}}\left(\sqrt{\left|1-\omega\right|} (\text{sgn}(\omega +1)+1)+\sqrt{\left| \omega +1\right|}(\text{sgn}(1-\omega )+1)\right)}{\sqrt{2\pi }e^{\frac{i\pi v}{2}} \sqrt{\left|1-\omega\right|}\sqrt{\left|\omega+1\right|}} \end{array} $$
The Fourier sine transform for the odd part is:
$$ \large\begin{array}{l} \sqrt\frac{2}{\pi}\int_{0}^{\infty}\frac{(-1)^{\frac{3}{4}}e^{-\frac{1}{2}i(\pi v+2x)}\left(\sqrt{-\frac{1}{x}}-e^{2ix}\sqrt{\frac{1}{x}}\right)\sin(\omega x)}{\sqrt{2\pi}}\,dx=\\ \frac{\left(-\frac{1}{2}+\frac{i}{2}\right) (-1)^{\frac{3}{4}} \left(\sqrt{\left|1-\omega \right|}(\text{sgn}(\omega+1)+1)-\sqrt{\left|\omega+1\right|} (\text{sgn}(1-\omega)+1)\right)}{\sqrt{2\pi}e^{\frac{i\pi v}{2}} \sqrt{\left|1-\omega\right|}\sqrt{\left|\omega+1\right|}} \end{array} $$