Proof: $ \int^{\infty}_{-\infty}\delta(t)^2 dt = 1 ??$ 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 ??$
 A: As in the other good answers and comments, the square of a distribution is not usually defined, as a distribution, or as anything else of a standard sort.
Nevertheless, having seen physicists routinely model very-short-range fields as "point scatterers", meaning the "potential" is a Dirac delta and we supposedly look at an operator $-\Delta+\delta$, this sort of question can be answered usefully in a less formal way (and without any hand-waving).
Namely, from $\int_{\mathbb R} e^{-\pi x^2}\;dx=1$, we have $\int_{\mathbb R} {e^{-\pi (x/\epsilon)^2}\over \epsilon}\;dx=1$. Further the functions $e^{-\pi(x/\epsilon)^2}/\epsilon$ converge to $\delta$ as $\epsilon\to 0$, in the topology on distributions. This is standard.
Unsurprisingly, the integrals of squares blow up as $\epsilon\to 0$, because
$$
\int_{\mathbb R} \Big({e^{-\pi(x/\epsilon)^2}\over \epsilon}\Big)^2\;dx
\;=\; {1\over \epsilon} \int e^{-2\pi x^2}\;dx \;=\; {1\over \epsilon\cdot \sqrt{2}} \to +\infty
$$
So, in contexts where $\delta$ is really just an idealization, the original formally meaningless question has a possibly-informative answer in the spirit of that idealization.
A: 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, $\int_{-\infty}^\infty \delta(x)^2\,\mathrm{d}x=\infty$.
A: Multiplication is not generally defined for "generalized functions". There is no such thing as $\delta \cdot \delta$. And even if you use the naive approach where $\int_{-\infty}^\infty f(x)  \delta(x) \mathrm{d} x=f(0)$, you will end up with $\int_{-\infty}^\infty \delta(x)^2 \mathrm{d} x=\delta(0)=\infty$. 
A: The OP should look at this:
https://mathoverflow.net/questions/48067/is-square-of-delta-function-defined-somewhere
http://nlab.mathforge.org/nlab/show/distribution#multiplication_of_distributions_14
http://en.wikipedia.org/wiki/Colombeau_algebra
