# Proof of n-th Derivative of Delta Dirac is Even or Odd Function (depending on n)

I want to show that the $(2m)$-th derivative of the Dirac delta function is even, while the $(2m+1)$-th derivative of the Dirac delta function is odd. Is my reasoning correct?

The exact line (later on down) where I am unsure of correctness is the following:

$$\frac{\mathrm{d}f(-x)}{\mathrm{d}x} = f'(-x) \cdot (-1)$$

(via the chain rule), but later I set x = 0, so that results in:

$$f'(-x) \cdot (-1) |_{x=0} = (-1) \cdot f'(0)$$

Proof that $1$-st derivative of Dirac delta function is odd:

We know: $$\int_{-\infty}^{\infty} f(t) \delta^{(1)} (t) \, \mathrm{d}t = (-1) \cdot f^{(1)}(0)$$

Calculating the following, we wish to show the following is equal to $(-1)$ times the previous equation, which will imply that the first derivative of the Dirac delta function is odd (by comparing the terms inside the integral).

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

(via substitution by parts), and now using integration by parts,

$$\int_{-\infty}^{\infty} f(-t) \delta^{(1)}(t) \, \mathrm{d}t = f(-t)\delta(t)|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \delta(t) \mathrm{d} f(-t)$$

Using the first and second equations in my post, we get:

$$f(-t)\delta(t)|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \delta(t) \mathrm{d} f(-t) = (-1) \cdot (-1) \cdot f'(0) = f^{(1)}(0)$$

(since the Dirac delta function and its derivatives, evaluated at any point other than 0, is equal to 0). This means the first derivative of the Dirac delta function is odd.

The first two equations in my post can be extended to higher derivatives of the function $f$:

$$\frac{\mathrm{d}f^{(n-1)}(-x)}{\mathrm{d}x} = f^{(n)}(-x) \cdot (-1)$$

$$f^{(n)}(-x) \cdot (-1) |_{x=0} = (-1) \cdot f^{(n)}(0)$$

And using similar reasoning as the first derivative case, the $(2m)$-th derivative of the Dirac delta function can be proved to be even, and the $(2m+1)$-th derivative of the Dirac delta function can be proved to be odd.

Specifically, are the last two (or first two) equations of my post valid?

You cannot treat distributions as ordinary functions. Instead, a distribution is odd if it vanishes on all even test functions. And that you've already shown in one of your first equations (involving the derivative of a test function at $0$). – WimC Jul 15 '12 at 21:56
Hi WimC: Thank you for your explanation. I believe the property can be proved using the following lines: 1. $\int_{-A}^{A} f(x) g(x) \, \mathrm{d}x = 0$ if $f(x)$ is even and $g(x)$ is odd. $f^{(2m+1)}(0) = 0$ if $f(x)$ is even. $f^{(2m)} = 0$ if $f(x)$ is odd. $\int_{-\infty}^{\infty} f(t) \delta^{(n)}(t) \, \mathrm{d}t = (-1)^n f^{(n)}(0)$. Suppose n is even and $f(t)$ is odd. Then $f^{(n)}(0) = 0$ and $\delta^{(n)}(t)$ is even for even $n$. Now, suppose n is odd and $f(t)$ is even. Then $f^{(n)}(0) = 0$ and $\delta^{(n)}(t)$ is odd for odd $n$. – jrand Jul 15 '12 at 22:57