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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 ??$

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    $\begingroup$ 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. $\endgroup$ Jul 31, 2013 at 9:55
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    $\begingroup$ We are dealing with distributions! $\endgroup$ Jul 31, 2013 at 9:59
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    $\begingroup$ 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? $\endgroup$ Jul 31, 2013 at 10:20
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    $\begingroup$ @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? $\endgroup$
    – Did
    Jul 31, 2013 at 12:54
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    $\begingroup$ 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. $\endgroup$ Jul 31, 2013 at 20:49

4 Answers 4

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

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  • $\begingroup$ 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?? $\endgroup$
    – kaka
    Jul 31, 2013 at 9:53
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    $\begingroup$ @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. $\endgroup$
    – Rhys
    Jul 31, 2013 at 10:57
  • $\begingroup$ @Rhys Because my voltage is assumed to be theoretically high amplitude spike. $\endgroup$
    – kaka
    Jul 31, 2013 at 13:16
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    $\begingroup$ @kaka In that case, you should be content with $\infty$ as the answer. $\endgroup$
    – Tunococ
    Aug 1, 2013 at 0:55
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    $\begingroup$ We can't really talk about $\delta(0)$. It is not even the pointwise limit of all approximations to $\delta$ as seen using $\phi_n(x)=2\pi n^3x^2e^{-\pi n^2x^2}$. $\endgroup$
    – robjohn
    Aug 1, 2013 at 1:24
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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

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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.

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

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    $\begingroup$ 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. $\endgroup$ Aug 1, 2013 at 1:42
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    $\begingroup$ @BabyDragon: the square of a function is that function multiplied by itself. $\endgroup$
    – robjohn
    Aug 1, 2013 at 2:12
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    $\begingroup$ 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. $\endgroup$ Aug 1, 2013 at 2:20
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    $\begingroup$ 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. $\endgroup$
    – robjohn
    Aug 1, 2013 at 2:25
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    $\begingroup$ @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. $\endgroup$
    – robjohn
    Aug 1, 2013 at 5:21

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