2
$\begingroup$

Let's say I have a dirac delta function:

$$\delta(x) = \begin{cases}\infty & x = 0 \\ 0 & x \ne 0\end{cases}$$

according to wikipedia, the Dirac delta function has the following property:

$$\delta(ax) = \frac{\delta(x)}{|a|}$$

(see, https://en.wikipedia.org/wiki/Dirac_delta_function#Scaling_and_symmetry)

So I attempt to prove that this property is true:

$$I = \int \limits_{-\infty}^{\infty} \delta(ax)~dx$$

let $u = ax$

$$du = a~ dx$$

$$dx = \frac{1}{a} du$$

integral becomes:

$$I = \int \limits_{-\infty}^{\infty} \frac{1}{a}\delta(u)~du$$

$$I = \frac{1}{a}$$

therefore:

$$\delta(ax) = \frac{1}{a}$$

My question is this, where does the absolute value comes from on the Wikipedia version of the property?

Example, I get this:

$$\delta(ax) = \frac{1}{a}$$

Wikipedia says this:

$$\delta(ax) = \frac{\delta(x)}{|a|}$$

$\endgroup$
3
  • 1
    $\begingroup$ What happens with the boundaries of the integral if you substitute $u=ax$ with negative $a$? $\endgroup$
    – Martin R
    Apr 27, 2020 at 16:34
  • $\begingroup$ i'm familiar with the calculus rule: $\int \limits_{x_1}^{x_0} f(x)~dx = - \int \limits_{x_0}^{x_1} f(x)~dx$ ... but, i'm not following you... if a<0... its just part of the constant "a"... why would you need to pull the negative sign out of a... $\endgroup$
    – Pico99
    Apr 27, 2020 at 16:42
  • $\begingroup$ This property is not derived, a linear change of variables is defined for singular distributions to coincide with what u-substitution gives for regular distributions. $\endgroup$
    – Maxim
    May 22, 2020 at 13:37

3 Answers 3

4
$\begingroup$

Note that for $a<0$, the bounds of integral are also reversed, so you will have $$I=\int_{\infty}^{-\infty} {1\over a}\delta(u)du=-{1\over a}$$

$\endgroup$
1
$\begingroup$

The delta function is clearly even, since for any $x \neq 0$, $\delta(x) = \delta(-x) = 0$. Since the delta function is even, we have that $\delta(ax) = \delta(-ax) = \delta(\vert a \vert x)$. Then, consider: \begin{align*} &\int \delta(\vert a \vert x)d(\vert a \vert x) = \ \ \ \ \ \ \ \ \ \ \ \text{(Let $u = \vert a \vert x$, so $du = \vert a \vert dx$)} \\ = &\int \delta(u)du = 1 = \int \delta(x)dx \\ &\int \delta(\vert a \vert x)d(\vert a \vert x) = \int \delta(x)dx \\ &\int \delta(\vert a \vert x)dx = \frac{1}{\vert a \vert}\int \delta(x)dx \end{align*} By the Fundamental Theorem of Calculus: \begin{align*} \frac{\mathrm{d}}{\mathrm{dx}}\int \delta(\vert a \vert x)dx &= \frac{1}{\vert a \vert}\frac{\mathrm{d}}{\mathrm{dx}}\int \delta(x)dx \\ \delta(\vert a \vert x) &= \frac{1}{\vert a \vert}\delta(x) \\ \delta(ax) &= \frac{1}{\vert a \vert}\delta(x) \end{align*}

$\endgroup$
0
1
$\begingroup$

lets assume $b > 0$.


$$I_1 = \int \limits_{-\infty}^{\infty} \delta(bx)~dx$$

let $u = bx$

$$du = b~ dx$$

$$dx = \frac{1}{b} du$$

$$u(x = \infty) = b x \big|_{x=\infty} = \infty$$

$$u(x = -\infty) = b x \big|_{x=-\infty} = -\infty$$

$$I_1 = \int \limits_{-\infty}^{\infty} \frac{1}{b}\delta(u)~du$$

$$I_1 = \frac{1}{b}$$


$$I_2 = \int \limits_{\infty}^{-\infty} \delta(-bx)~dx$$

let $u = -b~x$

$$du = -b~dx$$

$$dx = \frac{-1}{b}du$$

$$u(x = \infty) = -b x \big|_{x=\infty} = -\infty$$

$$u(x = -\infty) = -b x \big|_{x=-\infty} = \infty$$

$$I_2 = \int \limits_{-\infty}^{\infty} \frac{1}{-b} \delta(u)~du$$

$$I_2 = - \int \limits_{\infty}^{-\infty} \frac{1}{-b} \delta(u)~du$$

$$I_2 = \int \limits_{\infty}^{-\infty} \frac{1}{b} \delta(u)~du$$

$$I_2 = \frac{1}{b}$$


$$I_1 = I_2$$

$$\int \limits_{\infty}^{-\infty} \delta(-bx)~dx = \int \limits_{\infty}^{-\infty} \delta(bx)~dx = \frac{1}{b}$$

Now if we let "a" equal either "-b" or "b", then:

$$\delta(ax) = \frac{1}{|a|}$$

(the proceeding is true because all of the area for the dirac delta occurs when x=0.)

Now for the sake of completeness, also consider the case where a = 0:

$$\delta(0\cdot x) = \delta(0) = \frac{1}{|0|} = \infty$$

thus,

$$\delta(ax) = \frac{1}{|a|}$$

is true for all real values of a.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.