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I thought delta function's three properties.

$\delta(0)=\infty$, $\displaystyle\int\delta(t)dt=1$, and $\delta(t)=\delta(-t)$.

Therefore, if I do scaling by $a$, integral's value will be $1/|a|$.

$$ \therefore\delta(at)=\frac1{|a|}\delta(t) $$

In turn, differentiate both parts with t.

$$ \frac{d}{dt}\bigl[\delta(at)\bigr]=\frac{d}{dt}\left[\frac1{|a|}\delta(t)\right] $$

$$ \Leftrightarrow\ \ \ \delta'(at)=\frac1{|a|}\delta'(t) $$

Is this correct?

Danny_Kim
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    I am not going to answer, but it will be much easier to do whatever you want with the $\delta$- "function" if you use the real definition of it as a functional over the set $C_0^\infty (\mathbb R^n)$. Then you will not have anything to wander. See this link http://math.stackexchange.com/questions/1447887/dirac-delta-function-and-lebesgue-measurability/1447946#1447946 and the book that I recommended in it. – Svetoslav Sep 30 '15 at 09:25
  • This exact question was answered there. Why didn't you ask for explanations on this other page of yours? – Did Sep 30 '15 at 09:55
  • @Did The reason why I asked is the other page's answer. robjohn and you taught me that $\delta'(ax)=\frac1{a|a|}\delta'(x)$. So, I asked again for detailed answer. – Danny_Kim Sep 30 '15 at 10:04
  • I thought they are different question. The other page is about integral equation, and this page is not about integral. So, I posted new. Sorry~ – Danny_Kim Sep 30 '15 at 10:06
  • @Svetoslav thank you for link, I will read. – Danny_Kim Sep 30 '15 at 10:07

1 Answers1

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No: $\frac{d}{dt}(\delta(at)) = a \delta'(at)$. The chain rule (suitably interpreted) still applies.

Whether to consider it correct or not after this modification, depends on how formal you want to be in your treatment of distributions. (And since you say things like $\delta(0) = \infty$, I'm guessing you don't want to be very formal.)

mrf
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  • sorry, I cannot understand. $\delta(at)=\frac1{|a|}\delta(t)$ is correct. So, $\frac{d}{dt}\left[\delta(at)=\frac1{|a|}\delta(t)\right]$ is also correct, isn't it? Then, $\frac1{|a|}$ is constant. Therefore, $\frac1{|a|}\frac{d}{dt}\left[\delta(t)\right]$. I don't know where is incorrect. T.T – Danny_Kim Sep 30 '15 at 10:12
  • Oh my god!!!! I got it... I understand during comment.. $$ \delta'(at)=\frac{1}{a|a|}\delta'(t) $$ Thank you very much. – Danny_Kim Sep 30 '15 at 10:14