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I've seen following identity before $$ \int f(x)\delta(g(x))dx=\sum_i\frac{f(x_i)}{|g'(x_i)|} \tag{1} $$ Where $x_i$ are the roots to $g(x)$, but I've also seen cases where it has been written as $$ \int f(x)\delta(g(x))dx=\frac{f(x)}{|g'(x)|} \tag{2} $$ Where you don't take the roots into account. (e.g check out equation (23) here)

Now to my question: is (2) some kind of special case of (1)?

Edit: Added a link to an example.

Example: Suppose I have the following equation $$ \int x \delta(x+\sqrt{x^2+y^2+m^2-2xy\cos\theta}-y-m)dx $$ Where both $y,m$ are assumed to be real and positive.

Hence $f(x)=x$ and $g(x)=x+\sqrt{x^2+y^2+a^2-2xy\cos\theta}-y-a$ which has a root at $$ x_0=\frac{my}{m+y-y\cos\theta}=\frac{y}{1+\dfrac{y}{m}(1-\cos\theta)} $$ The derivative of $g(x)$ is $$ g'(x)=1+\dfrac{x-y\cos\theta}{\sqrt{x^2+y^2+m^2-2xy\cos\theta}} $$ Is it equally valid to just write that $$ \begin{align} \int x \delta(x+\sqrt{x^2+y^2+m^2-2xy\cos\theta}-y-m)dx=\frac{x}{1+\dfrac{x-y\cos\theta}{\sqrt{x^2+y^2+m^2-2xy\cos\theta}}} \end{align} $$ which would correspond to equation (2), as it is to write $$ \begin{align} \int x \delta(x+\sqrt{x^2+y^2+m^2-2xy\cos\theta}-y-m)dx= \frac{y}{1+\dfrac{y}{m}(1-\cos\theta)} \frac{1}{1+\dfrac{\dfrac{y}{1+\dfrac{y}{m}(1-\cos\theta)}-y\cos\theta}{\sqrt{\Big(\dfrac{y}{1+\dfrac{y}{m}(1-\cos\theta)}\Big)^2+y^2+m^2-2\Big(\dfrac{y}{1+\dfrac{y}{m}(1-\cos\theta)}\Big)y\cos\theta}}} \end{align} $$ Which would correspond to using equation (1)?

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    $\begingroup$ Where did you find the identity (2)? I may be wrong, but sifting property of delta makes the integrals into discrete values, leading me to think that (1) might be the right identity. $\endgroup$
    – Srini
    Commented Jan 12, 2018 at 16:12
  • $\begingroup$ I imagine the second author is being sloppy and either assuming $g$ only has one root, or assuming the unspecified domain of integration only includes one root, and calling it $x.$ Under those assumptions and notational abuses, yes, (2) is a special case of (1). $\endgroup$
    – ziggurism
    Commented Jan 12, 2018 at 16:13
  • $\begingroup$ Yes in the cases I've seen it $g(x)$ only has one root. $\endgroup$ Commented Jan 12, 2018 at 16:18

1 Answer 1

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You are integrating over $x$, so by definition your result cannot depend on $x$ anymore, it can only depend on some constant, say $x_0$.

This, your equation $1$ is correct, and $2$ is incorrect - it should be $x_0$ instead of $x$, and there is an implicit assumption that there is only one root in the domain of integration.

Your results from your example, are both incorrect for the same reason - they should not include $x$, just $x_0$. If you only have a single root, the corrected equation $2$ and equation $1$ are of course identical.

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  • $\begingroup$ It's no more wrong to write $\int \delta(g(x))\,dx=1/g'(x)$ than it is to write $\int^x\sin x\,dx=-\cos x.$ Which is to say, they're both bad notations, and you might consider them wrong, but some people definitely write that way. $\endgroup$
    – ziggurism
    Commented Jan 13, 2018 at 3:47
  • $\begingroup$ @ziggurism - It's a lot worse in this case. In your examples your stating the antiderivative, which is at least a function. Here, there is absolutely no meaning to the antiderivative, only to the definite integral, which is a constant, not a function. $\endgroup$ Commented Jan 13, 2018 at 3:58
  • $\begingroup$ It certainly led OP to think it should be a function of $x$, which is not so good $\endgroup$
    – ziggurism
    Commented Jan 13, 2018 at 4:45

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