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)?

  • 1
    $\begingroup$ 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 Jan 12 '18 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 Jan 12 '18 at 16:13
  • $\begingroup$ Yes in the cases I've seen it $g(x)$ only has one root. $\endgroup$ – Turbotanten Jan 12 '18 at 16:18

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.

  • $\begingroup$ 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 Jan 13 '18 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. $\endgroup$ – nbubis Jan 13 '18 at 3:58
  • $\begingroup$ It certainly led OP to think it should be a function of $x$, which is not so good $\endgroup$ – ziggurism Jan 13 '18 at 4:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.