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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – reuns
    Jul 13, 2022 at 5:05


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