Identity for rescaled Dirac Delta, $\delta(kx)$ I´m trying to proof the following Statement. 
$$\delta(kx)=\frac{1}{|k|} \delta(x).$$
I already tried to proof  and I got this.
$$u=kx \Rightarrow x=\frac{u}{k},dx=\frac{1}{k} du \\
\int_{-\infty}^\infty f(x) \delta(kx) dx = \int_{-\infty}^\infty f \left ( \frac{u}{k} \right ) \delta(u) \frac{du}{k}.$$ 
But I dont know how to proceed from here.
I know that I must catch the value of u. But my doubt is what value?
 A: When $T$ is a distribution and $k$ is a positive number, how should we define the rescaled distribution $T_k$? When $T$ is represented by a function $g$, we just want $g(kx)$. In terms of the integral against a test function $f$, this gives
$$
\int_{\mathbb{R}^n} g(kx) f(x) \,dx =k^{-n} \int_{\mathbb{R}^n} g(y) f(x/y) \,dx 
$$
So, for a general distribution $T$ we define  the rescaled distribution $T_k$ as 
$$
T_k(f)  = k^{-n} T(f(x/k))
$$ 
In the specific case of Dirac delta at $0$, the expression $k^{-n} T(f(x/k))$ evaluates to $ k^{-n}f(0)$. (You work in one dimension, $n=1$.)  
A: I worked this out referring to what you have already done and what Ian suggested. Also refer to https://proofwiki.org/wiki/Scaling_Property_of_Dirac_Delta_Function .
To prove $\delta(kx)=\frac{1}{|k|} \delta(x)$ , instead prove ${|k|}\delta(kx)= \delta(x)$, where k is a nonzero real constant. 
Use the definition of $\delta(x)$ consiting of the first condition:
$$\delta \left({x}\right) = \begin{cases}
+\infty & : x = 0 \\
0 & : x \neq 0
\end{cases}$$
and the second condition:
$$I=\int_{-\infty}^{+\infty} \delta \left({x}\right) dx = 1$$
Then show $\delta \left({kx}\right)$ satifies the second condition
$$I=\int_{-\infty}^{+\infty}|k| \delta \left({kx}\right) dx $$
$$=|k|\int_{-\infty}^{+\infty} \delta \left({kx}\right) dx = 1$$ 
since the first condition is obviously satisfied by $\delta \left({kx}\right)$ which is equal to $\infty$ for $x=0$ and is otherwise equal to zero.
Proceeding to find the integrals, $I$: let $y=kx$, then $dy=kdx$ and $dx=\frac{dy}{k}$.
So 
$$I=|k|\int_{-\infty}^{+\infty} \delta \left({kx}\right) dx
=|k|\int_{-\infty}^{+\infty} \delta \left({y}\right) \frac{dy}{k}$$
for $k>0$  $$I=\frac{|k|}{k}\int_{-\infty}^{+\infty} \delta \left({y}\right) dy =\frac{k}{k}\int_{-\infty}^{+\infty} \delta \left({y}\right) dy =1$$
and for $k<0$, noting that $y=kx$ will change the signs of the limits of integration, and also that reversing the limits multiplies the integrand by $-1$  $$I=\frac{|k|}{k}\int_{+\infty}^{-\infty} \delta \left({y}\right) dy =\frac{-|k|}{k}\int_{-\infty}^{+\infty} \delta \left({y}\right) dy =\frac{k}{k}\int_{-\infty}^{+\infty} \delta \left({y}\right) dy =1$$
which finishes the proof.
