Interpretation of the Dirac-measure property

First and foremost, apologies in advance for using an abuse of notation by placing the Dirac measure inside an integral. But given the circumstances, I have no choice.

This is essentially a word by word copy of an interpretation given on page 1 of these Berkeley notes:

The important property of the delta function is the following relation $$\displaystyle\int f(t) \delta(t) \, \mathrm{d}t = f(0)$$ for any function $f(t)$. This is easy to see. First of all, $\delta(t)$ vanishes everywhere except when $t = 0$. Therefore, it does not matter what values the function $f(t)$ takes except at $t = 0$. You can then say $f(t)\delta(t) = f(0)\delta(t)$. Then $f(0)$ can be pulled outside the integral because it does not depend on $t$, and you obtain the r.h.s.

Here's the problem, it was my understanding that $$\delta(t) = \begin{cases} 0 & \space \mathrm{for} \space t \ne 0 \\\infty&\ \mathrm{for} \space t = 0 \end{cases} $$ So by my logic this means that $\delta(0)=\infty$ and therefore undefined; which implies that when $t=0$ $$\displaystyle\int f(0) \delta(0) \, \mathrm{d}t = \displaystyle\int f(0) ~\infty \, \mathrm{d}t$$ which is manifestly not true and certainly not equal to $f(0)$.

Clearly I am missing the point of this argument, so if someone would be kind enough to explain it to me I would be most grateful.

Answer

A distribution is not a function, it is a functional acting on (a suitable space of) functions.

In particular, let's consider the functions of rapid decrease $\mathscr{S}(\mathbb{R}^d)$. Its topological dual, $\mathscr{S}'(\mathbb{R}^d)$, is the space of continuous linear functionals of $\mathscr{S}$, and it is called the space of tempered distributions.

How does a distribution work? It is a map that associates to each $f\in\mathscr{S}$, a complex number. Its action is usually denoted by $(\phi,\cdot)$, where $\phi\in\mathscr{S}'$. The Dirac delta distribution $\delta\in\mathscr{S}'$, is the distribution defined by: $$(\delta,f)=f(0) \quad, \quad f\in\mathscr{S}(\mathbb{R}^d)\; .$$

Now the rapid decrease functions are dense in the tempered distributions, in a suitable topology (the $\sigma(\mathscr{S}',\mathscr{S})$ one). And there is a natural identification of $f\in\mathscr{S}$ with the corresponding element $\tilde{f}\in\mathscr{S}'$: $\tilde{f}$ is the distribution whose action is defined by $$(\tilde{f},g)=\int_{\mathbb{R}^d}f(x)g(x)dx\quad ,\quad g\in\mathscr{S}(\mathbb{R}^d)\; .$$ So if the $\delta$ distribution was a real function $\delta(x)$ (but it is not!), we could write $$(\delta,f)=\int_{\mathbb{R}^d}\delta(x)f(x)dx=f(0)\; .$$ This is what it is usually done by physicists, but it is an abuse of notation. Another abuse of notation, is to write $\int_{\mathbb{R}^d}\delta(x)dx=1$. In principle, it is not possible to define the integral of a distribution. However this abuse of notation may be justified as follows:

• It is possible to approximate the delta distribution by rapid decrease functions, since $\mathscr{S}$ is dense in $\mathscr{S}'$. Indeed, given an integrable function $\eta$, such that $\int\eta(x)dx=1$, then $$\delta_h(x)=\frac{1}{h^d}\eta\bigl(\tfrac{x}{h}\bigr)$$ is an approximation of $\delta$, in the sense that $$\lim_{h\to 0}\int_{\mathbb{R}^d}\delta_h(x)f(x)dx=f(0)\quad ,\quad f\in\mathscr{S}(\mathbb{R}^d)\; .$$ Since, in addition, $$\int_{\mathbb{R}^d}\delta_h(x)dx=1$$ uniformly in $h$, it is tempting to conclude that $\int_{\mathbb{R}^d}\delta(x)dx=1$ (but it is not true, since- very roughly speaking -you are not allowed to take the limit inside the integral).

The above type of approximations $\delta_h$, also suggest the "pictorial" representation of $\delta$ as a function (but it is not!) that is zero everywhere, and infinity in zero. However, this is just a pictorial representation (that may be useful, but is not rigorous); so it cannot be used to argue pro or against the notation $\int_{\mathbb{R}^d}\delta(x)f(x)dx$ (that again is just an abusive notation, and not a rigorously defined integral).