# Cauchy's Integral Formula and Delta Functions

As we know from complex analysis, Cauchy's integral formula states:

$f(z_o)=\frac{1}{2\pi i}\int_\gamma{\frac{f(z)}{z-z_o}dz}$ for a closed contour $\gamma$.

However there is also the result from other areas of maths that states:

$f(x_o)=\int_R{f(x)\delta(x-x_o)dx}$ for some region $R$

Given the similarity in the results, is there any significance in comparing $\delta(x-x_o)$ to $\frac{1}{2 \pi i (x-x_o)}$ since basically integrating a function over them always gives the function's value at $x_o$

(Apologies in advance if this question may seem silly but I haven't found any satisfactory answers in my search so I thought I'd ask)

I'm not really an expert on $\delta$ functions... I don't really know anything about them at all, except that allegedly they have infinite 'mass' at one point, so this is really an extended comment. Nevertheless, it is a theorem that if a function is holomorphic on a region, and you have a contour in that region, you may deform that contour to any other contour as long as it stays in the region. If $f$ is holomorphic, then $\frac{f(z)}{z-z_0}$ is holomorphic everywhere except at $z_0$. So you have some contour containing the point $z_0$, you can think of this as contracting the loop 'all the way' to the point, so that you may as well be integrating over just the point. This seems reasonable since at the point $z_0$, the integrand blows up and so somehow should also have 'infinite mass' there.
• $\frac{1}{x}$ is the Hilbert transform of $\delta(x)$. – reuns Dec 18 '17 at 8:30