What is the complex Fourier transform of a winding number? My question regards complex and Fourier analysis.
If the winding number of a closed curve $\gamma$ in the complex plane with respect to $z_0=x_0+iy_0$ is given as follows:
$$\text{ind}_\gamma(z_0) := \frac 1{2\pi\mathrm i} \int_\gamma \frac{\mathrm d\zeta}{\zeta - z_0}.$$
What is it's complex Fourier transform?
I tried to evaluate the following, but don't really know what to do apart from swapping some factors in and out:
$$\widehat{\text{ind}}_\gamma(\omega_0)=\int_{-\infty}^{+\infty}\frac 1{2\pi\mathrm i} \int_\gamma \frac{\mathrm d\zeta}{\zeta - \omega_0} e^{-it\omega_0}dt.$$
Here, $\omega_0=\omega_x+i\omega_y$ is a complex number (does that even make sense for a frequency?). I am very thankful for any help on this one.
Edit: I came across this problem in an article by Pan et al. called "Sampling Curves with Finite Rate of Innovation". There it occurs in section II. A. in a slightly more general way, with a function in the numerator. I replaced that function by the constant function $1$ and ended up with a winding number.
 A: In the paper, a function with 2 real variables is defined
$$
I(x,y)=\frac1{2\pi i}\int_C\frac{F(u,v)}{u+iv-(x+iy)}d(u+iv).
$$
Of this function, the Fourier transform in both variables is considered, it is assumed that the integration over the (compact) curve $C$ can be moved out of the Fourier integration
\begin{align}
\hat I(ω_x,ω_y)&=A_{Fourier}\int_{\Bbb R^2}\frac1{2\pi i}\int_C\frac{F(u,v)}{u+iv-(x+iy)}\,d(u+iv)\,e^{-i(xω_x+yω_y)}\,d(x,y)
\\
&=\frac{A_{Fourier}}{2\pi i}\int_CF(u,v)\int_{\Bbb R^2}\frac{e^{-i(xω_x+yω_y)}}{u+iv-(x+iy)}\,d(x,y)\,d(u+iv)
\\
&=\frac{A_{Fourier}}{2\pi i}\int_CF(u,v)e^{-i(uω_x+vω_y)}\int_{\Bbb R^2}\frac{e^{i((u-x)ω_x+(v-y)ω_y)}}{(u-x)+i(v-y)}\,d(x,y)\,d(u+iv)
\\
&=\frac1{2\pi i}\int_CF(u,v)e^{-i(uω_x+vω_y)}\,d(u+iv)\cdot{A_{Fourier}}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{x+iy}\,d(x,y)
\end{align}
For the last factor one now finds
\begin{align}
A_{Fourier}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{x+iy}\,d(x,y)
&={A_{Fourier}}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}(ω_x-iω_y)}{(xω_x+yω_y)+i(yω_x-xω_y)}\,d(x,y)
\\
&=A_{Fourier}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{(xω_x+yω_y)+i(yω_x-xω_y)}\,\frac{d(xω_x+yω_y,yω_x-xω_y)}{ω_x+iω_y}
\\
&=\frac{A_{Fourier}}{ω_x+iω_y}\int_{\Bbb R^2}\frac{e^{ix}}{x+iy}\,d(x,y)
=\frac{A_{Fourier}}{ω_x+iω_y}\int_{\Bbb R^2}\frac{ix\sin(x)-iy\cos(x)}{x^2+y^2}\,d(x,y)
\end{align}
and somehow the last singular integral can be assigned a finite value.
