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In my impression it should be $1$ (or maybe, 2) but Wolfram Mathematica returns $0$. Who is wrong?

In general, in my impression $\int_a^b \delta(f(x))|f'(x)|dx$ should return the number of roots of the function $f(x)$ on the interval.

Anixx
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    What you have is ill defined because $\delta(f(x))$ is not always a defined quantity. It is only defined when $f(x)$ doesn't have a doubled (or higher) root - in other words $f$ and $f'$ cannot share roots. – Ninad Munshi Nov 28 '22 at 15:21
  • @NinadMunshi can it be reasonably generalized? Should it be $1$ or $2$ more reasonably? – Anixx Nov 28 '22 at 15:25
  • For $\epsilon>0$,$$\int_{-\infty}^\infty|x|\delta(x^2-\epsilon^2)dx=\frac{1}{2\epsilon}\sum_\pm\int_{-\infty}^\infty|x|\delta(x\mp\epsilon)dx=\frac{1}{2\epsilon}\sum_\pm|\pm\epsilon|=1.$$Hence $f(c):=\int_{-\infty}^\infty|x|\delta(x^2-c)dx$ is $1$ if $c>0$, but $0$ if $c<0$. So a "reasonably generalized" treatment of $f(0)$ has to make $f$ discontinuous at $c=0$. How do you propose to do that? – J.G. Nov 28 '22 at 15:33
  • @J.G. what the sum with plus-minus mean? – Anixx Nov 28 '22 at 15:35
  • If two quantities are convenient to label $a_+,,a_-$, $\sum_\pm a_\pm$ means $a_++a_-$. – J.G. Nov 28 '22 at 15:36
  • @J.G. so, it seems Mathematica gave one value and I expected the other. Hence I think it is better generalized to be $1$. – Anixx Nov 28 '22 at 15:38
  • How about this? $I=\int_{-\infty }^{\infty }|x| \delta \left(x^2\right) , dx = 2 \int_{0}^{\infty }x \delta \left(x^2\right) , dx$ Then $u=x^2,du=2xdx$ so $I= \int_{0}^{\infty } \delta \left(u\right) , du = \frac12 \int_{-\infty}^{\infty } \delta \left(u\right) , du=\frac12$. – Blitzer Nov 28 '22 at 15:42
  • @Blitzer Laplace transform of $\delta(x)$ is $1$, so this integral $\int_{0}^{\infty } \delta \left(u\right) , du$ is $1$ according to Laplace transform. – Anixx Nov 28 '22 at 15:48
  • @Blitzer $\int_0^\infty\delta(u)du$ is undefined because different values can be motivated depending on the choice of nascent delta. This is similar to the $f$ discontinuity I mentioned. – J.G. Nov 28 '22 at 15:50
  • @J.G. but the Laplace transform of $\delta(x)$ exists in the tables! – Anixx Nov 28 '22 at 15:52
  • About that... (Again, note the similarity to my $f$ example.) – J.G. Nov 28 '22 at 15:54
  • @J.G. and Annix - I know I was playing fast and loose in my comment and I acknowledge that my understanding is pretty light. In my head, $\delta(x)$ is a symmetrical sharp peak around the origin with an area under the curve of 1 and a limit. That's often enough to get me by - but not today! – Blitzer Nov 28 '22 at 22:12

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Let $f(x)=|x|$, $g(x)=x^2$, and $\delta_n(x)$ be a nascent Dirac Delta. We wish to determine the existence of the functional $\langle \delta\circ g,f\rangle$.

We now proceed to show that such a functional fails to exist. Note that we can write

$$\begin{align} \lim_{n\to \infty}\langle \delta_n\circ g,f\rangle&=\lim_{n\to \infty}\int_{-\infty}^\infty |x|\delta_n(x^2)\,dx\\\\ &=\lim_{n\to \infty}2\int_{0}^\infty x\delta_n(x^2)\,dx\\\\ &=\lim_{n\to \infty}\int_{0}^\infty \delta_n(x)\,dx\tag1 \end{align}$$

If $\delta_n(x)=n(H(x)-H(x-1/n))$, then $\lim_{n\to \infty}\langle \delta_n\circ g,f\rangle=1$.

However, if $\delta_n(x)=n(H(x+1/2n)-H(x-1/2n))$, then $\lim_{n\to \infty}\langle \delta_n\circ g,f\rangle=1/2$

Inasmuch as the value of the limit depends on the nascent Dirac Delta, the functional of interest fails to exist.

Mark Viola
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