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The Kronecker Delta can be written as the integral of the complex function $$f(n,z)=\frac{1}{2\pi i} z^{n-1} \ ,$$ where $n\in \mathbb{Z}$ and $z\in\mathbb{C}$ on a closed path $\mathcal{C}$ enclosing the origin $$ \delta_{n,0}= \oint_\mathcal{C} f(n,z) dz \ .$$ This can be seen as a trivial consequence of the Cauchy integral theorem $$ \delta_{n,0}= \oint_\mathcal{C} dz \frac{1}{2\pi i} z^{n-1} = \mathrm{Res}_{z=0}(z^{n-1}) \ .$$ The same function computed for $n=i\eta$ with $\eta\in\mathbb{R}$ and integrated on the real axis

$$ \int_0^\infty dy \, f(i \eta,y) =\int_0^\infty dy \frac{1}{2\pi i} y^{i\eta-1} = \int_{-\infty}^\infty dx \frac{1}{2\pi i} e^{i \eta x} = -i \delta(\eta) \ \ .$$

My question is the following. Is there a master function

$$ G(n) \equiv \int_\gamma dz f(n,z) $$

defined as the integral on some path $\gamma$ such that $G(n)\propto \delta_{n,0}$ and $G(i \eta)\propto \delta(\eta)$?

What are the path and the proportionality constants?

I do not know if it is useful but I noticed that $z^{n-1}$ has a pole also at infinity.

Pietro Dona
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    What should $G(0)$ be? – pregunton Jan 01 '21 at 18:14
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    That is an excellent question. If you approach 0 from the real axis, I would say one, while if you approach it from the imaginary axis, I would say that it is not defined or infinite. I am just wondering if it makes sense at all. I am not interested in G's existence in the whole complex plane but only on the real and imaginary axis. – Pietro Dona Jan 02 '21 at 19:05
  • Assuming that all the zeros of $G(z)$ are at the integer points of the real and imaginary axis, you can use Weierstrass Factorization to find an entire function which is the 'analytic continuation' of the Kronecker delta in the way you prescribed. I do not know, however, if this gives you the integral representation you're looking for. – VECH Jan 08 '21 at 13:37

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