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:

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

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


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

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:

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

  • 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


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


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*}


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:

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


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

is true for all real values of a.


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