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. – Mhenni Benghorbal Jul 31 '13 at 9:55
• 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? – kjetil b halvorsen Jul 31 '13 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 '13 at 12:54
• @kaka Do you plan to leave the present question in disarray because people suggest that you stick to mathematics and you do not like the news, just like you did with this other one? – Did Jul 31 '13 at 15:22
• 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. – Christian Blatter Jul 31 '13 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 '13 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 '13 at 10:57
• @Rhys Because my voltage is assumed to be theoretically high amplitude spike. – kaka Jul 31 '13 at 13:16
• @kaka In that case, you should be content with $\infty$ as the answer. – Tunococ Aug 1 '13 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 '13 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. – Baby Dragon Aug 1 '13 at 1:42
• @BabyDragon: the square of a function is that function multiplied by itself. – robjohn Aug 1 '13 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. – Baby Dragon Aug 1 '13 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 '13 at 2:25
• would the downvoter care to comment? – robjohn Jan 12 '15 at 21:04