# Kramers-Kronig relations for a Gaussian function

Consider a function of a complex variable $$\omega$$ which is given by $$f(\omega) = e^{-\omega^2/2}$$.

This function is symmetric, holomorphic everywhere, and vanishes as $$|\omega| \rightarrow \infty$$. Thus the Kramers-Kronig relations tell us that, given the real part we can find the imaginary part:

$$\Im[f(\omega)] = \frac{2\omega}{\pi} \mathcal{P} \int_{0}^{\infty} d\omega' \frac{\Re[f(\omega')]}{\omega'^2 - \omega^2}$$

Here $$\mathcal{P}\int$$ represents the Cauchy principal value.

We expect of course to find that $$\Im[f(\omega)] = 0$$ for real $$\omega$$, but Mathematica seems to disagree. Attempting to solve this numerically, we have

f = Exp[-x^2/2]
g[y_] := (2 y)/Pi * NIntegrate[-(Re[f]/(x^2 - y^2)), {x, 0, y, Infinity}, Method -> "PrincipalValue"]
Plot[{g[y], Im[f /. x -> y]}, {y, 0, 5}]


we obtain an output

which clearly is not zero everywhere. I'm inclined to think that this a problem with how I've written my Mathematica code, but I really can't work my head around it. Mathematica doesn't complain about convergence, the function satisfies all the constraints (I think!) and I'm fairly certain I've specified the integral correctly (i.e. giving the excluded singular point in the integration region).

What's going on?

I'm not sure that PSE is the best site to ask this question, but I'll answer it anyway. The main issue is that the KK relations do not apply on $$f$$. Intuitively, the Fourier transform of $$f$$ is again a gaussian so it is not causal.
From a purely spectral point of view, the issue is that you only looked at the properties of $$f$$ on the real line. You need to look at the upper half complex plane. You'll notice that it does not decay to zero when $$\omega\to \pm i\infty$$. In fact you have an essential singularity there.