Inverse function with Dirac Delta

Question

We know that the inverse function of

$ y=\log(x) $ is $ y =\exp(x) $.

However, what would be the inverse of

$ y=\log(x)+ \sum_{n=1}^{\infty}\delta (x-n) $?

I have tried with Mathematica, and whenever there is a Dirac Delta, for example at the point $ x=1 $, at this point the computer gives a 'null' value, there is a white point on the screen at this point $ x=1 $. Very curious.

Answer 1

The delta function, viewed as a distribution, takes a (sufficiently well-behaved) "test" function as its input, and spits out the value of the function at a certain point as its output. For example, let $\delta_a$ be the delta distribution centered at $a$, then for any real-valued test function $f$ on the real line, one obtains $\delta_a(f) = f(a)$. Notice that if $g\neq f$ is any other function that has the same value as $f$ at $a$, then we also have $\delta_a(g) = g(a)$ so that

$\delta_a(f)=\delta_a(g)$.

This means that $\delta_a$ is not one-to-one and therefore it is not invertible. For the function you gave, it is essentially the same story.