# Inverse function with Dirac Delta

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.

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Phew...I believe it is ill-defined even if you approximate the delta function in the usual ways (a peak and then take the limit as it gets infinitely sharp at constant unit area) because that function is not monotonic. – dmckee Jan 25 at 21:46
i guess the inverse function of $log(x) +\sum_{n=1}^{\infty} \delta (x-n)$ is just $y=exp(x)$ except for the points $n=1,2,3,4,5,6,....$ where is ill-defined. – Jose Garcia Jan 25 at 22:24
Jose, I put some questions on your deleted question "Fourier transform (inverse) of a Heavside function" at the meta site and I hope you don't mind. Cheers... – draks ... Jan 27 at 16:35
it does not matter draks :) regards.. – Jose Garcia Jan 27 at 20:11

## migrated from physics.stackexchange.comJan 26 at 3:02

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.