Intuition for why $\int_{-\infty}^{\infty} f(t) \delta(t) dt = f(0)$

Question

What is a good, intuitive, 1st-year Calculus student, layman's-terms reason that the Dirac delta function $\delta(t)$ is defined to have the characteristic $\int_{-\infty}^{\infty} f(t) \delta(t) dt = f(0)$?

The best I've got right now is that since the integral really only takes place when t = 0 (because it's 0 everywhere else due to $\delta(t)$ being 0 everywhere else), and it appears that $\int_0^0 \delta (t) dt = 1$ (weird but I'll allow it), then $f(t) \delta(t)$ just scales $\delta(t)$ by $f(t)$ at the point when t = 0...aka it's just scaling by $f(0)$. Thus if $\int_0^0 \delta (t) dt = 1$, then it would make sense that multiplying this integral by the scalar $f(0)$ would just give the result $\int_0^0 f(0)\delta (t) dt = f(0)$, and thus (kinda) $\int_{-\infty}^{\infty} f(t) \delta(t) dt = f(0)$

Is my intuition acceptable? Or is there a better way to think about it?

Answer 1

A personally better picture would be as follows:

1. We always fix a very, very small unit $\mathrm{d}x$ of change, with respect to which all the other changes are measured. For example,

   $$ \frac{\mathrm{d}f(x)}{\mathrm{d}x} \approx \frac{f(x+\mathrm{d}x) - f(x)}{\mathrm{d}x} \qquad \text{and}\qquad \int_{a}^{b} f(x) \, \mathrm{d}x \approx \sum_{a + k\mathrm{d}x \in [a, b]} f(a+k\mathrm{d}x) \, \mathrm{d}x, $$

   etc.

2. $\delta(x)$ is a "function" that assigns a unit mass near $x = 0$. For example, this may be realized as

   $$ \delta(x) \approx \begin{cases} \frac{1}{\mathrm{d}x}, & |x| < \mathrm{d}x / 2, \\ 0, & \text{otherwise}, \end{cases} $$

   Then we may argue that

   $$ \int_{-\infty}^{\infty} f(x) \delta(x) \, \mathrm{d}x \approx \sum_{k} f(k\mathrm{d}x) \, \delta(x) \, \mathrm{d}x \approx f(0). $$

Answer 2

The Dirac delta "function" isn't a function, and the things you write in your question do not make literal sense. Instead, it is an abuse of notation that has a certain plausibility to it, but which most people who use it don't understand. The actual picture is this:

We have a natural pairing between (nice function), $\langle f,g\rangle =\int_{\mathbb R} f(x)g(x)dx$, and if $f$ is continuous and $g$ is only non-zero on a very small interval around $0$, then because continuous functions don't vary a lot on very small intervals, $\int_{\mathbb R} f(x)g(x)dx\approx f(0)\int_{\mathbb R}g(x)dx$ As the support of $g$ gets smaller and smaller, the approximation becomes better and better.

So, for example, we could set $$g_n=\begin{cases}2n \quad & \text{if }x\in(\frac{-1}{n},\frac{1}{n})\\ 0 \quad &\text{otherwise}\end{cases}$$ and $\lim_{n\to\infty}\langle f,g_n\rangle=f(0)$ whenever $f$ is continuous. But there is no literal function that the $g(x)$ converge to, and even if it did, we wouldn't necessarily have that the integral commutes with the limit. But we do have that as an operator on continuous functions, the pairings are approaching the operator that sends $f(x)$ to $f(0)$. So we say "wouldn't it be nice if there were a function that we were integrating against to make this happen?" And then we call that non-existent function $\delta(x)$.