# Integral involving Dirac delta: two different results?

I am evaluating the integral over all space

$$\int \delta \left(r^2 - R^2\right) d \vec r$$

At first, I did this:

$$\int \delta \left(r^2 - R^2\right) d \vec r = 4 \pi \int_0^\infty \delta \left(r^2 - R^2\right) r^2 dr = 4 \pi R^2$$

But then someone made me notice that we can use the property

$$\delta[f(r)] = \sum_i \frac{\delta (r-r_i)}{\mid f'(r_i) \mid}$$

where $r_i$ are the roots of $f$. The only root we have to consider, since $r\geq0$, is $+R$; I thus obtained

$$\delta \left(r^2 - R^2\right) = \frac{\delta(r-R)}{2R}$$

which holds the result

$$\int \delta \left(r^2 - R^2\right) d \vec r = \frac{4 \pi}{2R} \int \delta(r-R) r^2 dr = 2 \pi R$$

Which result is the right one?

Update

I am starting to think that the first result is wrong because I am basically assuming that

$$\int_0^\infty \delta[f(x)] g(x) dx = g(x_i)$$

where $x_i$ is the only root of $f(x)$ contained in the interval $[0,\infty)$. But this is wrong because the normalization ($\mid f'(x_i) \mid^{-1}$) is missing...

• @5xum You are right. There was also -I think- an error in my question. I have updated it. – valerio Jul 18 '16 at 9:02
• @5xum But isn't that also true for, say, $\int_{-\infty}^{\infty} \delta(x) dx = 1$? – valerio Jul 18 '16 at 9:10
• @5xum $\int_{-\infty}^{\infty} \delta(x) dx$ has no bounds and nothing I am integrating over, but it is not ill-defined. – valerio Jul 18 '16 at 9:12
• @5xum Integral over all space means over $\mathbb{R}^3$, if I'm not mistaken. – pregunton Jul 18 '16 at 9:13
• Ohhh, I missed the part "integral over all space". Dang. Sorry about that... – 5xum Jul 18 '16 at 9:13

## 1 Answer

Suppose you make the substitutions $u=r^2$, $U=R^2$. Then $du=2r\,dr$ and so $$4\pi \int_0^\infty \delta(r^2-R^2)\,r^2\,dr =4\pi \int_0^\infty \delta(u-U)\,\frac{\sqrt{u}}{2}\,du=2\pi \sqrt{U}=2\pi R$$ in accordance with the normalized formula. So a substitution which linearizes the argument of the delta reduces the integral to the standard case.