# Dirac delta function composition

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)?

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)?

• 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. – Srini Jan 12 '18 at 16:12
• 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). – ziggurism Jan 12 '18 at 16:13
• Yes in the cases I've seen it $g(x)$ only has one root. – 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.
• 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. – ziggurism Jan 13 '18 at 3:47
• It certainly led OP to think it should be a function of $x$, which is not so good – ziggurism Jan 13 '18 at 4:45