TITLE: Proof: $ \int^{\infty}_{-\infty}\delta(t)^2 dt = 1 ??$
QUESTION [6 upvotes]: Let $\delta(t)$ be the Dirac-Delta function. I know that its area is 1, and amplitude is $\infty$. 
Then, how to prove that:
$ \int^{\infty}_{-\infty}\delta(t)^2 dt = 1 ??$

REPLY [7 votes]: Consider the approximations to $\delta$ given by $\phi_n(x)=ne^{-\pi n^2x^2}$ :
$$
\begin{align}
\int_{-\infty}^\infty \phi_n(x)^2\,\mathrm{d}x
&=\int_{-\infty}^\infty n^2e^{-2\pi n^2x^2}\,\mathrm{d}x\\
&=\frac{n}{\sqrt2}\int_{-\infty}^\infty e^{-\pi x^2}\,\mathrm{d}x\\
&=\frac{n}{\sqrt2}
\end{align}
$$
As $n\to\infty$, $\int_{-\infty}^\infty \phi_n(x)^2\,\mathrm{d}x\to\infty$. Therefore, in the sense of distributions, $\int_{-\infty}^\infty \delta(x)^2\,\mathrm{d}x=\infty$.