# 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:

$\int^{\infty}_{-\infty}\delta(t)^2 dt = 1 ??$

• By the way, where did you find that this integral equals to 1? I gave you a hint to evaluate the integral but you won't find this answer. Jul 31, 2013 at 9:55
• We are dealing with distributions! Jul 31, 2013 at 9:59
• I cannot see any way to define this integral as being equal to 1, and many ways to evaluate it to "+ infinty", that is, it is divergent. Can you give us the reference saying it is =1? Jul 31, 2013 at 10:20
• @kaka The WP page you referred to is mediocre (to put it politely) and not even self-consistent (note that the paragraphs "Energy" and "Convolution" about the Dirac function are contradictory). Why not try a mathematical reference?
– Did
Jul 31, 2013 at 12:54
• We are told that $1+1=3$ and that we either should prove this or should be able to refute this by producing a contradiction. The first is clearly impossible in the "normal" mathematical world, and the second requires a recursion to the absolute basics of our science. Jul 31, 2013 at 20:49

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 $\int_{-\infty}^\infty \delta(x)^2 \mathrm{d} x=\delta(0)=\infty$.

• Suppose $\delta(t)$ is a voltage signal, and we are interested in computing the energy or power of this signal then how you will go??
– kaka
Jul 31, 2013 at 9:53
• @kaka Why would you ever want to model a voltage signal with a delta function? The power itself might be approximated by a delta function.
– Rhys
Jul 31, 2013 at 10:57
• @Rhys Because my voltage is assumed to be theoretically high amplitude spike.
– kaka
Jul 31, 2013 at 13:16
• @kaka In that case, you should be content with $\infty$ as the answer. Aug 1, 2013 at 0:55
• We can't really talk about $\delta(0)$. It is not even the pointwise limit of all approximations to $\delta$ as seen using $\phi_n(x)=2\pi n^3x^2e^{-\pi n^2x^2}$.
– robjohn
Aug 1, 2013 at 1:24

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.

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, $\int_{-\infty}^\infty \delta(x)^2\,\mathrm{d}x=\infty$.

• What if we let $\phi_n=\begin{cases} -n^2|x-\frac{1}{n}|+n &&& 0<x<\frac{2}{n}\\ 0 &&& \mbox{otherwise}\\ \end{cases}$ and $\theta_n=\begin{cases} -n^2|x+\frac{1}{n}|+n &&& -\frac{2}{n}<x<0\\ 0 &&& \mbox{otherwise}\\ \end{cases}$. Both of these approach the delta function, but when we multiply them and integrate, we get zero. Aug 1, 2013 at 1:42
• @BabyDragon: the square of a function is that function multiplied by itself.
– robjohn
Aug 1, 2013 at 2:12
• The sequence of functions, $\phi_n$ approaches $\delta$. So does $\theta_n$. This means that $\phi_n\theta_n$ should approach this thing called $\delta^2$. Now when we integrate $\phi_n\theta_n$ this should approach the integral of $delta^2$. But when we compute the square this way we get zero. Aug 1, 2013 at 2:20
• This is simply not true. $\int_{-\infty}^\infty|\phi_n(x)-\theta_n(x)|\,\mathrm{d}x=2$. The two do not approach each other.
– robjohn
Aug 1, 2013 at 2:25
• @BabyDragon: There are many approximations of $\delta$, but as you've seen, the product of two different approximations does not necessarily behave like the square of a single approximation. The action of a convergent sequence of distribution approximations is measured against a fixed function. The action measured against a changing sequence of functions is unpredictable.
– robjohn
Aug 1, 2013 at 5:21