# Powers of the Dirac Delta "Function"

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

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

$$\blacksquare$$

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

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 $$+\infty$$.