A delta function has the sampling property, it picks out the value of a function $f(x)$ at a point $a$ $$ f(a) = \int_{-\infty}^{\infty} \delta(x - a) f(x) dx $$ The Fourier integral theorem says $$ f(a)=\frac{1}{2\pi} \int_{-\infty}^\infty e^{-ipa}\left(\int_{-\infty}^\infty e^{ipx }f(x)\,d x \right) \,dp =\frac{1}{2\pi} \int_{-\infty}^\infty \left(\int_{-\infty}^\infty e^{-ip(x-a)} \,dp \right)f(x)\,d x $$ so the "Fourier" delta function can be defined as $\delta(x-a)=\frac{1}{2\pi} \int_{-\infty}^\infty e^{-ip(x-a)} $.
However the sampling property is also true for Cauchy's integral formula $$ f(a) = \frac{1}{2\pi i} \oint_{\partial D} \frac{f(z)\,dz}{z-a} $$ So the "Cauchy" delta function is $\delta(z-a) = \frac{1}{2\pi i} \frac{1}{z-a}$.
So this looks different. However is it really different? It is often possible to use contour integration and Cauchy's integral formula to compute Fourier integrals . So I was wondering: is there a direct connection between both representations of the delta function, if one chooses an appropriate contour to compute the contour in Cauchy's integral formula so that basically Cauchy's integral formula becomes equal to the Fourier integral theorem?