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It seems to me that $$ \int_{-\infty}^\infty \int_{-\infty}^{\infty} \delta(x^2 + y^2 - R^2) dx dy $$ should evaluate to $2\pi R$, the perimeter of a circle of radius $R$, but I'm having trouble getting this answer. I tried using the identity $$ \delta(x^2 + y^2 - R^2) = \frac{1}{2\sqrt{R^2 - y^2}}\left[\delta\left(x - \sqrt{R^2 - y^2}\right) + \delta\left(x + \sqrt{R^2 - y^2} \right)\right] $$ But it wasn't clear to me how to evaluate the $x$ integral after making this substitution.

How do you evaluate the above integral? Does the integral represent the perimeter of a circle of radius $R$? If not, how should such an integral be written?

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  • $\begingroup$ It's like integrating $1$ along the circle, a set of Lebesgue's measure $0$. $\endgroup$ – Jakobian Jun 28 at 8:30
  • $\begingroup$ It doesn't evaluate to $2\pi R$, that's why it doesn't work out for you. It would if you had $\delta(\sqrt{x^2+y^2}-R)$, but in this case, your delta has a little extra stretched because its argument is some function, not just an identity in terms of $x-x_0=0$. $\endgroup$ – orion Jun 28 at 9:50
  • $\begingroup$ See this answer. You can verify that both of your formulas will give the same answer (if you're using the second one, just integrate $\delta(x - x_0)$ wrt $x$), but the answer is $\pi$, not $2 \pi R$. $\endgroup$ – Maxim Jun 28 at 23:02
  • $\begingroup$ @Maxim that linked answer leaves more to be desired, as the assumptions on the functions in question are left unspecified, as well as the passage from the substitution rule to an actual definition of the associated integral operator (I believe there is a question related to function inversion along the way). $\endgroup$ – pre-kidney Jun 29 at 4:01
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Keep in mind the dirac delta is not a function, but rather a distribution. In fact, its precise definition (in this case) is that for all test functions $f(x,y)$, $$ \int_{\mathbb R^2}f(x,y)\delta(x^2+y^2-R^2)\ dx\ dy :=\int_{\theta=0}^{2\pi}f(R\cos\theta,R\sin\theta)\ d(R\theta). $$ The left side has no independent meaning - it is defined to equal the right side.

In your case, take $f(x,y)$ to be the constant $1$ function and you get your answer.

EDIT: After the discussion in the comments, I see this question really boils down to asking

What is a precise definition of $\delta(g(x,y))$?

More generally, one may ask for the definition of $\delta(g)$ where $\delta$ is the $n$-dimensional delta function and $g\colon \mathbb R^n\to \mathbb R^n$ is a smooth function. And the answer turns out to be that there is no such definition that has all the desired properties in general, even for $n=1$. An explicit counterexample is furnished in Remark 2 of these notes by Terry Tao.

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  • $\begingroup$ I guess I have a more general question, which I might ask separately. How do you evaluate integrals of the form $\int_{\mathbb{R}^2} f(x,y) \delta(g(x,y)) dxdy$? $\endgroup$ – Charles Hudgins Jun 28 at 9:25
  • $\begingroup$ By parametrizing the region $g(x,y)=0$ (it will have one fewer variable) and integrating over that. $\endgroup$ – pre-kidney Jun 28 at 9:26
  • $\begingroup$ Is there no way to write it in terms of $g$ and maybe its derivatives? $\endgroup$ – Charles Hudgins Jun 28 at 9:27
  • $\begingroup$ No, not in general. Also, note that your identity is equivalent to using a different parametrization of the circle (by splitting into two halves and integrating each separately). $\endgroup$ – pre-kidney Jun 28 at 9:28
  • $\begingroup$ That's disappointing. Just to clarify, if $\gamma : [0,1] \to \mathbb{R}^2$ is a parametrization of $g(x,y)$, then $\int f(x,y)\delta(g(x,y)) = \int_0^1 f(\gamma(t)) |\gamma'(t) | dt$. Is that right? $\endgroup$ – Charles Hudgins Jun 28 at 9:37
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Delta functions are just shorthand notation to restricting integration to a lower dimensional region (in this case, a circle). Delta function only makes real sense inside an integral, and has its own rules for changing variables.

In this case, just change variables to polar coordinates (not forgetting the Jacobian):

$$\int_{-\infty}^\infty\int_{-\infty}^\infty \delta(x^2+y^2-R^2)dx\,dy= \int_0^{2\pi}\int_0^\infty \delta(r^2-R^2)r\,dr\,d\phi$$ The $\phi$ integral is trivial: $$\color{red}{=2\pi\int_0^\infty \delta(r^2-R^2)r\, dr }$$

Consider now the defining property of the delta function that says it just evaluates the function at its peak position:

$$\int f(x)\delta(x-x_0)dx=f(x_0)$$

The other important part is the variable substitution rule:

$$\delta(g(x))=|g'(x_0)|^{-1}\delta(x-x_0)$$ where $x_0$ is the zero of $g(x)$.

In our case, $g(r)=r^2-R^2=(r-R)(r+R)$ has a zero at $r=R$ (the negative one is outside the integration range). We have $g'(R)=2R$. So, the above integral in $\color{red}{\text{red}}$ can be transformed into

$$=2\pi\int_0^\infty \frac{1}{2R}\delta(r-R)r\, dr=\pi$$

where we took into account that this delta function just evaluates the function $\frac{r}{2R}$ at $r=R$.

Notice that it is extremely important that the argument of the delta function is not $\sqrt{x^2+y^2}-R=r-R$ but $x^2+y^2-R^2=r^2-R^2$.

Another way of looking at this is to start from the red integral again, but change variables to $r^2=u$.

$$2\pi\int_0^\infty \delta(r^2-R^2)r\, dr= 2\pi\int_0^\infty \delta(u-R^2)\frac{du}{2}=\pi $$ where the delta function now just evaluated function $\frac{1}{2}$ at $u=R^2$ which doesn't matter because the function was just a constant.

The moral of the story: it's not just important where the zero of the delta function's argument is, you have to take into account the functional form (especially the derivative). Stretched delta isn't the same as delta even if the peak is at the same place. If you imagine it as a narrow and tall Gaussian peak, you can see that stretching the function by changing variables changes its area.

You could also write your delta function as $$\delta(r^2-R^2)=\delta((r-R)(r+R))\equiv \delta(2R(r-R))\neq \delta(r-R)$$ where I took into account that the pre-factor $r+R$ is only important where $r=R$, so I put that in.

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  • $\begingroup$ Actually, the "defining property" of the "delta function" you have given is valid only for the $1$-dimensional case. You have not specified the meaning of your multidimensional "delta function", and in fact the entire question centers are the precise definition of the $2$-dimensional case. $\endgroup$ – pre-kidney Jun 28 at 18:41
  • $\begingroup$ In fact, your answer is showing that there are several sensible integral operators that one can associate to a given "delta expression", and thus it pays to be precise when one is working with multivariable delta and simply say what is the integral operator one has in mind for a given expression. $\endgroup$ – pre-kidney Jun 28 at 18:46
  • $\begingroup$ @pre-kidney $\delta(t) = \lim_{n \to \infty} n \ 1_{t \in [0,1/n]}$ yields $$\iint_\Bbb{R^2} \delta(x^2+y^2-R^2)dxdy =\lim_{n \to \infty} n \iint_\Bbb{R^2} 1_{x^2+y^2 -R^2\in [0,0+1/n]} dxdy= \lim_{n \to \infty} n ( \pi (R^2+1/n)-\pi R^2) = \pi$$ $\endgroup$ – reuns Jun 29 at 0:46
  • $\begingroup$ @reuns I appreciate the sentiment but there are still some holes. Limit I assume is in the sense of distributions (otherwise just wrong), but then it is applied with a function substitution - this operation is not well-defined for distributions and arbitrary functions, and even still it is unclear why this operation would be continuous wrt the weak topology. Possible to cook up nonsense counterexamples that nevertheless follow the same reasoning you have presented... $\endgroup$ – pre-kidney Jun 29 at 4:05
  • $\begingroup$ @pre-kidney This is a definition only then you check in what sense it is (or it is not) well-defined. In most case if it is we can interpret the limit as a continuous linear functional on some space. Here $\varphi \mapsto \iint \varphi(x,y)\delta(x^2+y^2-R^2)dxdy$ is continuous $C^\infty(\Bbb{R}^2) \to \Bbb{R}$ and $n1_{x^2+y^2-R^2 \in [0,1/n]}$ converges to $\delta(x^2+y^2-R^2)$ in the space of compactly supported distributions as well as in the dual of $C^0(\Bbb{R})$ $\endgroup$ – reuns Jun 29 at 15:02

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