We know that the product of the delta function and another function samples the latter function. That is, $$ \delta(t-\tau)f(t)=\delta(t-\tau)f(\tau) $$

Does the doublet function retain this same property? That is, does the following hold true: $$ \delta'(t-\tau)f(t)=\delta'(t-\tau)f(\tau) $$ My reasoning is that $\delta'(t-\tau)$ is only nonzero at $\tau$, and therefore the value of $f$ at any time other than $\tau$ does not matter, in the same manner as the delta function product.

One of this issues I have with this result though is that computing the Laplace transform of the doublet function does not seem to work out. $$ \mathcal{L}\{\delta'(t)\} = \int_{-\infty}^{\infty} \delta'(t)e^{-st}dt = \int_{-\infty}^{\infty} \delta'(t)e^{-s\times0}dt = \int_{-\infty}^{\infty} \delta'(t)dt = 0 $$ The final integral is zero since the doublet function is odd. However, we know that since the doublet is the derivative of the delta, its Laplace transform must be $\mathcal{L}\{\delta'(t)\}=s$.

I would appreciate if someone could offer some insight.


2 Answers 2


Your first equation is correct. For derivatives of the Dirac delta impulse you get slightly more involved expressions. For $\delta'(t)$ the following holds:


Of course we assume that $f(t)$ and $f'(t)$ are continuous at $t=0$.

With $(1)$ you obtain the correct result for the Laplace transform of $\delta'(t)$:

$$\begin{align}\int_{-\infty}^{\infty}\delta'(t)e^{-st}dt&=\int_{-\infty}^{\infty}\big[\delta'(t)e^{-s\cdot 0}+se^{-s\cdot 0}\delta(t)\big]dt\\&=\int_{-\infty}^{\infty}\delta'(t)dt+s\int_{-\infty}^{\infty}\delta(t)dt\\&=s\end{align}$$

The general formula for the $n^{th}$ derivative of the Dirac delta impulse is [1]


[1] A. Papoulis, The Fourier Integral and Its Applications, p. 274.

  • $\begingroup$ Thank you for your answer. Can you explain where the minus sign comes from in equation (1)? At first I thought you were using the derivative product rule, but now I'm not sure. $\endgroup$ Feb 20, 2022 at 19:16
  • 1
    $\begingroup$ @LetterSized: You're right: $[f(t)\delta(t)]'=f'(t)\delta(t)+f(t)\delta'(t)$, and $[f(t)\delta(t)]'=f(0)\delta'(t)$, which results in Eq. $(1)$. $\endgroup$
    – Matt L.
    Feb 20, 2022 at 19:26
  • $\begingroup$ Excellent, thanks! $\endgroup$ Feb 20, 2022 at 19:27
  • $\begingroup$ "Your first equation is correct" well. I'd call it ok, maybe not correct $\endgroup$ Feb 20, 2022 at 21:57
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    $\begingroup$ @LaurentDuval: I would say it's perfectly correct if interpreted correctly. Of course, what it means is $$\int_{-\infty}^{\infty}f(t)\delta(t-\tau)\phi(t)dt=\int_{-\infty}^{\infty}f(\tau)\delta(t-\tau)\phi(t)dt$$ $\endgroup$
    – Matt L.
    Feb 21, 2022 at 10:11

First, the Dirac delta and its "derivatives" are not traditional functions, and we ought not to say that they are valued at $\tau$ with $-\infty$ or with $+\infty$ (or possessing both values).

Second, this "unit doublet" could however shape some intuition on what happens when we perform some operation (product, convolution, etc.) with them. And this can be used as a mnemonic. You can read for instance a couple of answers at What is the first derivative of Dirac delta function?

Third, let's not forget conditions. This kind of product $\delta\times f$ is not generic. It requires some conditions on $f$. Classically, $f$ should be a smooth enough function, unless you can get weird and incorrect results. For once, it is very difficult to make sense of an expression like $\delta^2$.

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    $\begingroup$ Even though $\delta(t)$ is not defined at $t=0$, we can certainly say that $\delta(t)=0$ for all $t\neq{}0$, no? Hence why (assuming $f$ is sufficiently smooth) the product of a delta (and its derivatives) with $f$ only depends on the behavior of $f$ at $t=0$. Though I agree with points 2 and 3. I could have better specified $f$, but there's a reason why I posted this question here and not the math stack exchange :). $\endgroup$ Feb 20, 2022 at 21:31
  • $\begingroup$ Saying something precise about a non-function is out of my grasp. It may depend on the test function space you choose. I however would be ok, in practice, to consider the Dirac ti be zero almost everywhere $\endgroup$ Feb 20, 2022 at 21:54

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