# Dirac $\delta(g(x))$ with complex roots $x_i$

Hey we all know about this infamous Wikipedia page Dirac-delta composition related to the Dirac-$\delta$ in composition with a function $g(x)$.

But I wonder if the roots of $g(x) = 0$ happens to be of the form $r+\mathrm{i}s$ with $r,s\in\mathbf{R}$; is $\int_{\Omega} f(x)\delta(g(x)) \mathrm{d}x =^?0$, here $\Omega$ is an integration path on the real line.

I suspect it should be zero since we never encounter them, but I would like to hear your arguments.

Thanks.

Perhaps you are concerned that something like the Residue Theorem might have an effect on such an integral. The Residue Theorem only holds for functions meromorphic on the domain inside the path of integration, and $\delta\circ g=0$ is not. $delta$ is zero everywhere but where its argument is $0$. Therefore, if the roots of $g$ are not real, then $\delta\circ g=0$ on $\mathbb{R}$.
In the terms of distributions, the support of $\delta\circ g$ is the set of zeros (the roots) of $g$, therefore, if $\Omega\cap\{z:g(z)=0\}=\{\}$, then $$\int_\Omega f(x)\,\delta\circ g(x)\,\mathrm{d}x=0$$
If the function $g(x)$ has no roots in the domain of Integration, the result should always be zero. This has nothing to do with composition but rather with the properties of the delta function - namely that it's equal to zero everywhere except for $\delta(0)$.