Fourier transform of Gaussian-like you have never seen Let $m$ be a positive integer, I would then like to compute the "Fourier transform" (but now this is an area integral over the entire complex plane)
$$\int_{\mathbb C} \overline{x}^m e^{-ikx} e^{-\vert x \vert^2/2}\ dx$$
here $k \in \mathbb R$ but unfortunately the integral is over $\mathbb C$ rather than $\mathbb R$. Is there still a way to compute it?
 A: Indeed, an exact closed form exists.
Define the factor $\kappa=-2ik$.
Then,
$$\begin{align}
I(k)&=\int_0^\infty\int_0^{2\pi} \bar z^m e^{-ikz}e^{-\left\lvert z \right\lvert^2 /2} d\theta \,rdr \\
&=\int_0^\infty\int_0^{2\pi} r^{m+1}e^{-im\theta} e^{\kappa re^{i\theta}/2}e^{-r^2 /2} d\theta dr \\
&=\int_0^\infty r^{m+1}e^{-r^2 /2}\left(\int_0^{2\pi}e^{-im\theta} e^{\kappa re^{i\theta}/2} d\theta\right) dr \\
&=\int_0^\infty r^{m+1}e^{-r^2 /2}\left(\oint_{|z|=1}z^{-m} e^{\kappa rz/2} \frac{dz}{iz}\right) dr \qquad (1)\\
&=\int_0^\infty r^{m+1}e^{-r^2 /2}\left(2\pi i\cdot\frac{(\kappa r/2)^m}{i\cdot m!}\right) dr \qquad (2)\\
&=\frac{2\pi(\kappa/2)^m}{m!}\int_0^\infty r^{2m+1}e^{-r^2 /2}dr \\
&=\frac{2\pi(\kappa/2)^m}{m!}\int_0^\infty 2^m u^me^{-u}du \qquad (3)\\
&=\frac{2\pi\kappa^m}{m!}\int_0^\infty u^me^{-u}du \\
&=\frac{2\pi\kappa^m}{m!}\cdot m! \qquad (4)\\
I(k)&=2\pi\kappa^m
\end{align}
$$
$(1)$: by the substitution $z=e^{i\theta}$.
$(2)$: by residue theorem.
$(3)$: by the substitution $u=\frac{r^2}{2}$.
$(4)$: by the definition of Gamma function $\displaystyle{\int_0^\infty x^n e^{-x}dx=\Gamma(n+1)=n!}$.
