In this question the multiplication of two impulse functions is discussed

However, what if I have two shifted impulses say:

$z(t) = \delta(t-\tau)\cdot \delta(t-\beta)$

Now if I want to integrate $z(t)$ how would I go about it? Let's say the integral is:

$$\int_{-\infty}^{\infty}\delta(t-\tau)\cdot \delta(t-\beta) dt $$

  • $\begingroup$ That vanishes for $\beta \neq \tau$ and is undefined otherwise. In general, if you encounter a product of two Dirac distributions with the same variable, you've most likely made a mistake elsewhere. It's not something that should happen. $\endgroup$
    – Jazzmaniac
    Mar 14, 2023 at 10:36
  • $\begingroup$ I know the result and have an intuitive idea for it. But can it be expressed as a Dirac Delta function? $\endgroup$ Mar 14, 2023 at 10:38
  • $\begingroup$ @Jazzmaniac: At first I thought exactly the same, but now I think that since convolution of two delta impulses IS well-defined, that integral is actually a convolution and the result should be $\delta(\beta-\tau)$. $\endgroup$
    – Matt L.
    Mar 14, 2023 at 10:39
  • $\begingroup$ @Matt L. How exactly is it a convolution? If it were shouldn't the integral have been: $\int_{-\infty}^{\infty}\delta(t-\tau)\cdot \delta(t) dt$ $\endgroup$ Mar 14, 2023 at 10:43
  • 1
    $\begingroup$ @Jazzmaniac: OK, I'll write up an answer and I'd be grateful if you pointed out my mistake. $\endgroup$
    – Matt L.
    Mar 14, 2023 at 10:52

1 Answer 1


Multiplication of two Dirac impulses is undefined. However, convolution of two Dirac impulses is defined [1]:


Writing out $(1)$ gives


where I've used the fact that $\delta(t)$ is even. The last integral in $(2)$ has the same form as the integral in the OP. Hence, using $(1)$ with $t_1=\tau$ and $t-t_2=\beta$ we obtain


As correctly pointed out by Jazzmaniac, the integral in $(3)$ is a distribution, and as such it only makes sense if wrapped by an integral with $\beta$ or $\tau$ as the integration variable. This is always the case with Dirac impulses and distributions in general. Eq. $(3)$ has to be understood in that sense.

[1] A. Papoulis, The Fourier Integral and Its Applications, p. 275

The quantity $\delta^2(t)$ is undefined, however, $$\small{\delta(t-t_1)*\delta(t-t_2)=\int_{-\infty}^\infty\delta(\tau-t_1)\delta(t-\tau-t_2)\,d\tau=\delta[t-(t_1+t_2)]\tag{I-29}}$$ as we can easily see from $\text{(I-15)}$.

The product of two distributions $g_1(t)$ and $g_2(t)$ is, in general, undefined; however, their convolution is given by $$\small{\begin{array}{rl}\displaystyle\int_{-\infty}^\infty\bigg[\int_{-\infty}^\infty g_1(\tau)g_2(t-\tau)\,d\tau&\hspace{-1em}\bigg]\phi(t)\,dt\\&\displaystyle\hspace{-1em}=\int_{-\infty}^\infty g_1(\tau)\bigg[\int_{-\infty}^\infty g_2(t-\tau)\phi(t)\,dt\bigg]d\tau\tag{I-15}\end{array}}$$ as we see, by a formal change of the order of integration.

  • $\begingroup$ I would agree if there was another integral wrapped around the whole thing, so that the result appeared under an integral, and that integral would have to be over either $\beta$ or $\tau$. On its own, without such context, the correct answer would be the one I have given above, i.e. it vanishes for $\beta\neq\tau$ and is undefined otherwise. $\endgroup$
    – Jazzmaniac
    Mar 14, 2023 at 11:10
  • $\begingroup$ In the strict sense, distributions don't exist outside of an integral. Or in other words, distributions represent linear forms and the integral is part of that form. $\endgroup$
    – Jazzmaniac
    Mar 14, 2023 at 11:12
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    $\begingroup$ @Jazzmaniac: Yes, I agree, but this is always the case with Dirac impulses (or distributions in general). They make sense under an integral. So I think we can agree that what Eq. (3) of my answer means is that IF in an appropriate environment (i.e., inside an appropriate integral), the result of the integral in the OP equals a Dirac impulse with argument $\beta-\tau$. $\endgroup$
    – Matt L.
    Mar 14, 2023 at 11:15
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    $\begingroup$ Good, then we're on the same page! $\endgroup$
    – Jazzmaniac
    Mar 14, 2023 at 11:16
  • 5
    $\begingroup$ I'm just glad to see Jazz hanging around. $\endgroup$ Mar 14, 2023 at 15:03

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