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I will present a sketch of a proof for the following by induction:

$$\int_{-\infty}^{\infty}\delta(x)^{n}f(x)\mathrm{d}x = {(-1)^n}f^{(n-1)}(0); n>1$$

I will also use the definition

$$\delta(x):=\Theta'(x)$$

Where

$$\Theta(x) = \begin{cases}1 \text{ if }x\geq0\\0\text{ otherwise}\end{cases}$$

I'm not really sure if the steps are right or even if the above is correct, so I need help verifying if this is valid and what kind of function $f$ could be.

Case n=1

$$\int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x = f(0) = (-1)^0f^{(0)}(0)$$

Case n+1

$$\int_{-\infty}^{\infty}\delta^{n+1}(x)f(x)\mathrm{d}x = \int_{-\infty}^{\infty}\left(\Theta'(x)\right)^{n+1}(x)f(x)\mathrm{d}x = \left(\Theta'(x)\right)^nf(x)\Bigg{|}_{-\infty}^{\infty} - \int_{-\infty}^{\infty}\left(\Theta'(x)\right)^{n}f^{(1)}(x)\mathrm{d}x$$

By definition of the delta function, $\Theta'(x)=0$ for $x\neq0$, so

$$\int_{-\infty}^{\infty}\delta^{n+1}(x)f(x)\mathrm{d}x = - \int_{-\infty}^{\infty}\left(\Theta'(x)\right)^{n}f^{(1)}(x)\mathrm{d}x = - \int_{-\infty}^{\infty}\delta(x)^nf^{(1)}(x)\mathrm{d}x$$

Using the hypothesis

$$-\int_{-\infty}^{\infty}\delta(x)^nf^{(1)}(x)\mathrm{d}x = -(-1)^{n}f^{(n)}(0) = (-1)^{n+1}f^{n}(0)$$

$\blacksquare$

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  • $\begingroup$ Hi Ícaro Lorran. Welcome to Math.SE. Did you mean the $n$th derivative (rather than the $n$th power) of $\delta(x)$? $\endgroup$
    – Qmechanic
    Commented Aug 13, 2023 at 10:44
  • $\begingroup$ Hello Qmechanic, I actually meant the nth power. But according to the last answer, this looks invalid $\endgroup$ Commented Aug 13, 2023 at 15:35
  • $\begingroup$ The posted question gives a heuristic development of the nth derivative of the Dirac Delta. $\endgroup$
    – Mark Viola
    Commented Aug 14, 2023 at 2:43

1 Answer 1

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This is incorrect. One can not work with powers of the Dirac delta function, as if it were a ordinary function. To see that such a calculation goes wrong, let us use the simplest representation of the delta function, namely a block. I.e. we define the delta function $\delta(x)$ as $1/\epsilon$ for $-\epsilon /2 < x < +\epsilon /2$ and $0$ elsewhere. We now get:

$$\int _{-\infty} ^{\infty} \delta(x)^n f(x)dx = \frac {1}{\epsilon ^n} \int_{-\epsilon/2} ^{\epsilon/2}f(x)dx = f(0) / \epsilon^{n-1}$$

As required we now take the limit of $\epsilon$ to $0$, and then we see that for all powers of $n$ larger than $1$ the result diverges, i.e. goes to $+\infty$.

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