I want to calculate $$f(q)=\int_{-\infty}^\infty G(r)p(r)e^{iqr} dr$$ with complex $q$.
I know the Fourier transform of $p(r)$ for complex $q$. I.e. $f_{p}(q)=\int_{-\infty}^\infty p(r)e^{iqr} dr$ is solved.
and G(r) is a Gaussian function $$G(r)=e^{-ar^2}$$ I want to use the convolution theorem to solve this problem and thus need the Fourier transform of $G(r)$. I know that for real $q$, the Fourier transform of a Gaussian is another Gaussian. My central question is: is this still valid if $q$ is complex?
Wikipedia seems to hint that the answer may not be trivial: https://www.wikiwand.com/en/Fourier_transform#/Complex_domain
"Depending on the properties of f, this might not converge off the real axis at all, or it might converge to a complex analytic function for all values of ξ = σ + iτ, or something in between"
and the explanation following that quote goes over my head. Despite searching I have yet to confirm this key detail.
So, please help me!: Is it true that the Fourier transform of a Gaussian function is still $$f_{g}(q)=\int_{-\infty}^\infty e^{-ar^2} e^{iqr} dr = \sqrt{\pi/a} e^{-\pi^2 q^2 /a}$$ for complex $q$? Thank you
Okay I think I figured this out with the help of this post:
Complex Frequency Shifting in Fourier Transform
I want to confirm that
$$f_{g}(q)=\int_{-\infty}^\infty e^{-ar^2} e^{iqr} dr = \sqrt{\pi/a} e^{-\pi^2 q^2 /a}$$
or otherwise find a similar solution, when $q$ is complex.
if $q=x+iy$ $$f_{g}(q)=\int_{-\infty}^\infty e^{-ar^2} e^{i(x+iy)r} dr$$ or $$f_{g}(q)=\int_{-\infty}^\infty e^{-ar^2} e^{-yr} e^{ixr} dr$$
Using the convolution theorem, this is equivalent to $$f_{g}(q)=(\int_{-\infty}^\infty e^{-ar^2} e^{ixr} dr) \circledast (\int_{-\infty}^\infty e^{-yr} e^{ixr} dr)$$ Which since the LHS is a Gaussian and the RHS is a plane wave is $$f_{g}(q)=((\sqrt{\pi/a} e^{-\pi^2 x^2 /a}) \circledast(c \delta(x-y)))(x)$$ where c is some constant that may or may not exist, and which I am currently too tired to properly think about.
Then because the delta function is kind of like the identity function of convolution (something I vaguely remember from DSP - apologies if that is not rigourously true...), the convolution just scales the Gaussian by $c$ and shifts it by $y$... so the answer is $$f_{g}(q)=c(\sqrt{\pi/a}) e^{-\pi^2 (x-y)^2 /a}$$ So the effect of complex q is just to scale and shift the resulting Gaussian. I'm not 100% confident, and I suspect there's a simpler answer. I'd appreciate anyone jumping in to correct me if so!