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

Question

$\delta$ stands for Dirac delta distribution. I thought I could just make a substitution and get what I want. However in 1 dimension I get: $$\int_{-\infty}^{\infty}\delta(ax)\varphi(x)dx=\frac{1}{a}\int_{-\infty}^{\infty} \delta(y)\varphi(y/a)dy=\frac{1}{a}\varphi(0)=\frac{1}{a}\int_{-\infty}^{\infty} \delta(x)\varphi(x)dx$$ Where I used substitution $y=ax$ and ${\mathrm d}y=a{\mathrm d}x$ You can see that the absolute value is missing.

I also tried with different definition: $$\delta(ax)=\int_{-\infty}^{\infty}e^{ikax}dk=\frac{1}{a}\int_{-\infty}^{\infty}e^{ilx}dl=\frac{1}{a}\delta(x)$$ Now I've made the same mistake twice. Can you please point out what am I doing wrong?

Answer 1

Suppose $a < 0$. Then $$ \int_{-\infty}^{+\infty} \delta(ax)\varphi(x) dx = \frac 1 a \int_{+\infty}^{-\infty} \delta(y) \varphi(y/a) dy = - \frac 1 a \int_{-\infty}^{+\infty} \delta(y) \varphi(y/a) dy$$ Pay attention to the limits on the integrals!