$\delta(ax) = \frac{\delta(x)}{|a|}$, where does absolute value come from? 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:
$$\delta(ax) = \frac{\delta(x)}{|a|}$$
 A: Note that for $a<0$, the bounds of integral are also reversed, so you will have $$I=\int_{\infty}^{-\infty} {1\over a}\delta(u)du=-{1\over a}$$
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*}
A: 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.
