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Take the following integral: $$\int^{\infty}_{-\infty}f(x)\delta(cx)dx$$ Why can I not say that the argument of $\delta(cx)$ "picks out" the value $x=0$, making the integral $f(0)$? By "picks out", I am trying to use the rule that we only care about the value in the range of the bounds that makes the argument zero.

I know that I can do a change of variable, so I actually get that the integral evaluates to $\frac{1}{c}f(0)$, but I am not sure what is wrong with the initial claim.

Relativisticcucumber
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    As you see it is a matter of definition. You can/should define $\int^{\infty}{-\infty}f(x)\delta(cx)dx$ as $\lim{n\to \infty} \int^{\infty}{-\infty}f(x) 2n , 1{|cx|<1/n},dx$ that is $f(0)/|c|$ when $f$ is continuous. This is really desirable to think to distributions as limits (in a certain topology) of sequences of functions. In particular it makes it compatible with usual changes of variable. – reuns Jul 13 '22 at 05:05

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