Can $e^{c\delta(t)}$ be rewritten some how?

Question

Can \begin{align*} e^{c\delta(t)} \end{align*} be rewritten some how? Where $\delta(t)$ is a delta function and $c$ is some constant.

Answer 1

The problem here is that $\delta$ isn't really a function--rigorously, it's a functional. The idea that $\int_{-\infty}^\infty f(t) \delta(t) dt=f(0)$ for any real function is perfectly fine, but the notion that $\delta$ is some bizarre function that assumes an infinitely tall value in an infinitesimally small region to have unit area can't be made rigorous. In other words, $\delta(t)$ is shouldn't be taken outside of an integral.

Therefore, it makes sense to call $\int_{-\infty}^\infty (\cdot)\delta(t) dt$ a functional: an operator that takes as input a real-valued function $f$ and outputs a real number $\int_{-\infty}^\infty f(t)\delta(t)dt = f(0)$. Let's write this in more suggestive notation: $\delta[f] := f(0)$.

Composing this functional with a function doesn't product a function, then. Let $\mathbb{R}^\mathbb{R}$ denote the set of real valued functions (generally the notation $Y^X$ is the set of functions from the set $X$ to the set $Y$). Then $\delta$ is a function from $\mathbb{R}^\mathbb{R}$ to the real numbers, $\mathbb{R}$. We denote this by $\delta: \mathbb{R}^\mathbb{R} \rightarrow \mathbb{R}$. Following this style, let's write $e^x$ instead as $\exp(x)$, andthen we can write $\exp: \mathbb{R} \rightarrow \mathbb{R}$. We can connect the two functions like this: $\exp \circ \delta:\mathbb{R}^\mathbb{R} \xrightarrow{\delta} \mathbb{R} \xrightarrow{\exp}\mathbb{R}$. The composite, then, is an operator that takes a function $f$ to the real number $e^{f(0)}$.

Now, for an operator $T$ that has the same domain and range, it is sometimes possible to define its exponential so as to produce a new operator via a power series: $e^T(f):= \sum\frac{T^n(f)}{n!}$. Here $T^n$ denotes the composition of $T$ with itself $n$ times [ex. $T^3(f) = T(T(T(f))) ]$. This doesn't work because $\delta$ takes functions to numbers, not other functions. $\delta(\delta(f))$ isn't a meaningful expression; it would evaluate to $\delta(f(0))$, which doesn't make sense.

For an example where this would work, let's consider the scaled derivative operator, $\alpha \frac{d}{dx}$ where $\alpha$ is a real number. This operator will act on smooth functions to produce new smooth functions, namely the derivative of the original function scaled by $\alpha$. Incredibly enough, using the above definition, $e^{\alpha \frac{d}{dx}}$ is a translation operator. Specifically, $(e^{\alpha \frac{d}{dx}}(f))(x) = \sum \frac{\alpha^n \frac{d^nf}{dx^n}}{n!} =f(x+\alpha)$. That's amazing!

For a proof of this (at least for analytic functions), see the second answer here:Exponential of the differential operator.

Summary: The delta "function" isn't really a function, but is best viewed as an operator that takes real-valued functions to real numbers by the equation $\delta[f] = f(0)$. Composing this operator with the exponential function would then yield a new operator that takes a function $f$ to the real number $e^{cf(0)}$. For operators with the same domain and range, you can use a power series to define a notion of exponentiating an operator to produce a new operator. That is not the case here, but it leads to some interesting results if you want to learn more.

Disclaimer: there are other ways to view the delta functional and other approaches to defining it's exponential that are probably more fruitful for further development