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Currently, I am studying the Laplace transforms. I know that $X(p) = p$ cannot match with any $x(t)$ function, since $X(p) \notin L^1$. However, I decided to try computing the reverse Laplace transform to see what it gives.

I use the following properties:

$$\delta (t) \overset{\mathcal{L}}{\longleftrightarrow} 1 \tag{1}$$

$$\frac{d}{dt}x_0(t) \overset{\mathcal{L}}{\longleftrightarrow} pX_0(p) \tag 2$$

Using $X_0(p) = 1$, I deduce:

$$\color{red}{\frac{d}{dt}\delta(t)} \overset{\mathcal{L}}{\longleftrightarrow} p \tag 3$$

That's why I wonder: what is the derivative of the Dirac impulse ? It is, perhaps, a "forbidden" generalised function?

Rodrigo de Azevedo
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    The derivative of the Dirac impulse is defined by the formal equation $\int_{-\infty}^\infty \delta'(x) f(x) dx = -f'(0)$. Look up the term "distributional derivative" for more on why that makes any sense. – Ian Dec 25 '22 at 14:00
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    See this, for example: https://dsp.stackexchange.com/questions/68732/what-is-the-first-derivative-of-dirac-delta-function – Enrico M. Dec 25 '22 at 14:03
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    The inverse Laplace transform of things like $p$ is exactly why Schwartz invented (tempered) distributions. – reuns Dec 25 '22 at 15:52
  • Yes, you either have differentiable delta function but having no square or exponentiable but having no derivative. The usual one is the first. https://math.stackexchange.com/questions/4627326/squarable-but-not-differentiable-delta-function – Anixx Jan 28 '23 at 00:20

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There is nothing special or "forbidden" about derivative of Dirac delta or any other generalized function. By definition, derivative $\varphi'$ of some generalized function $\varphi$ is a generalized function that satisfies $$ \left<\varphi', f\right> = \int_{\mathbb{R}} \varphi'(t) f(t)\mathrm{d}t = - \int_{\mathbb{R}} \varphi(t) f'(t)\mathrm{d}t = -\left<\varphi, f'\right> $$ for all test functions $f$.

Specifically, $\delta'$ is defined as $\left<\varphi', f\right> = -f'(0)$.

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