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I'm using the definition of the one dimensional dirac-delta function, $\delta(x)$, being, $$\int_{-\infty}^{+\infty}\delta(x)f(x)dx = f(0) \tag1$$ and I'm doing a question which asks me to (via ordinary integral manipulations) show that, for $a \in \mathbb{R}$, $$\delta(ax)=\frac{1}{|a|}\delta(x).\tag2$$ The solution said to use substitution of $u=|a|x$ and $du=|a|dx$ to get $$\int_{-\infty}^{+\infty}\delta(ax)f(x)dx = \frac{1}{|a|}\int_{-\infty}^{+\infty}\delta(\pm u)f\biggl(\frac{u}{|a|}\biggl) du= \frac{1}{|a|}f(0) \tag3.$$ And then by equating the LHS of $(3)$ with with the LHS of $(1)$ we get the identity in $(2)$. That makes sense. Only thing is I don't get why one would have used the substitution, $u=|a|x$, as opposed to the seemingly more natural, $u=ax.$ Can anybody tell me why they did that in the solution? The short answer is because that is what gets the correct answer. But the identity in $(2)$ seems to be a major/common one of the Dirac Delta function, which seems to imply that, $$\delta(ax)=\frac{1}{a}\delta(x),$$ doesn't hold for negative $a$. If that is the case can someone explain why?

I hope my question is clear.

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 1, 2018 at 18:53
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    $\begingroup$ What is $\delta(-x)$? Equivalently, what are the even and odd parts of $\delta(x)$? $\endgroup$
    – Sean Lake
    Feb 1, 2018 at 18:53
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    $\begingroup$ Hint: Trace what happens to the integration limits when $a<0$. $\endgroup$
    – Qmechanic
    Feb 1, 2018 at 18:55
  • $\begingroup$ @Qmechanic Ahh that makes sense, I got it thanks! $\endgroup$
    – Premez
    Feb 1, 2018 at 19:09

2 Answers 2

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Fisrt, let us come back to the initial step:

$$ \int_{-\infty}^{+\infty}dx\,\delta(ax)=\frac{1}{a}\int_{-\infty}^{+\infty}adx\,\delta(ax) $$

For $a$ positive we can make $du=adx$ and the limit of the integration are the same. For $a$ negative, the limits of the integration are interchanged when we perform the integration in $u$. If we interchange back the limits of integration this will cost an extra $(-1)$ factor, since:

$$ \int_{a}^{b}du=-\int_{b}^{a} du $$

Then we have that if $a$ is negative, we have $\delta(ax)=(-1)\frac{1}{a}\delta(x)$. At the end we can write for all signs of $a$: $\delta(ax)=\frac{1}{|a|}\delta(x)$

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Hint1 :

\begin{equation} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(\alpha x+\beta\right)\mathrm{h}\left(x\right)\mathrm dx =\int\limits_{\boldsymbol{-} \mathrm{sign}(\alpha) \, \infty}^{\boldsymbol{+}\mathrm{sign}(\alpha) \, \infty}\!\!\!\!\!\!\!\!\delta\left(u\right)\mathrm{h}\left(\dfrac{u-\beta}{\alpha}\right)\dfrac{\mathrm d u}{\alpha} \tag{01} \end{equation}


$^{1}$ Physical meaning of the Jacobian in relation to Dirac delta function, $\:\color{blue}{\textbf{Example A :}}$.

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