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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?

Bernard
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Charles Hudgins
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  • It's like integrating $1$ along the circle, a set of Lebesgue's measure $0$. – Jakobian Jun 28 '19 at 08:30
  • 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 $x-x_0=0$. – orion Jun 28 '19 at 09:50
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    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 $2 \pi R$. – Maxim Jun 28 '19 at 23:02
  • @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). – pre-kidney Jun 29 '19 at 04:01

3 Answers3

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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.

pre-kidney
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  • I guess I have a more general question, which I might ask separately. How do you evaluate integrals of the form $\int_{\mathbb{R}^2} f(x,y) \delta(g(x,y)) dxdy$? – Charles Hudgins Jun 28 '19 at 09:25
  • By parametrizing the region $g(x,y)=0$ (it will have one fewer variable) and integrating over that. – pre-kidney Jun 28 '19 at 09:26
  • Is there no way to write it in terms of $g$ and maybe its derivatives? – Charles Hudgins Jun 28 '19 at 09:27
  • 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). – pre-kidney Jun 28 '19 at 09:28
  • 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? – Charles Hudgins Jun 28 '19 at 09:37
  • Yes, unless you want to be pedantic, in which case there is an extra sign ambiguity coming from a choice of orientation of the curve $g(x,y)=0$. It is another reason why the multidimensional dirac delta notation is not extremely common in rigorous mathematics. More precisely, there are two natural ways to define what $\delta(g(x,y))$ means, depending on how you orient the region. (In the case of the circle, this is clockwise vs counterclockwise.) – pre-kidney Jun 28 '19 at 09:43
  • I would seem, based on orion's answer, that the set determined by $g(x,y) = 0$ is not all that matters. It seems like taking $g(x,y) = x^2 + y^2 - R^2$ yields a different answer than taking $g(x,y) = \sqrt{x^2 + y^2} - R$, even though their zero sets are identical. This contradicts the above integral expression. What's going on here? – Charles Hudgins Jun 28 '19 at 10:06
  • There is no contradiction, simply inadequacy of definitions. The other answer confuses the definition of the single variable and multivariate delta functions. In fact there is an inherent ambiguity in the multidimensional case, which can only be rectified if one writes down the precise integral operator they mean (instead of manipulating ill-defined functions). – pre-kidney Jun 28 '19 at 18:44
  • Ah. That's disappointing, but I suppose it is to be expected. I was reading a paper recently which argued that there really is no good way to define the dirac delta in many contexts. I didn't expect I would run into a problem of definition with such an elementary example. – Charles Hudgins Jun 28 '19 at 18:56
  • Remark 2 doesn't apply here, there are no stationary points on the curve. Postulating the change of variables property and the sifting property, like orion does in his answer, gives a consistent definition of $\delta(g)$. Parametrizing $g(x, y) = 0$ and evaluating the integral of $f(\gamma(t)) |\gamma'(t)|$ won't be compatible with this definition. – Maxim Jun 28 '19 at 23:08
  • @Maxim actually it does apply here, since the original $g(x,y)$ to which the question was applied does have stationary points. – pre-kidney Jun 28 '19 at 23:14
  • Stationary points on the curve $g = 0$ (aka critical points). It doesn't matter if $x^2 + y^2 - R^2$ has a stationary point elsewhere, the definition I'm talking about remains consistent. – Maxim Jun 28 '19 at 23:28
  • Please feel free to answer this question and provide your definition, I will be especially pleased if it is rigorous and includes all hypotheses. However, the point still stands that there is no such definition in general and one must make suitable assumptions on the function one is composing your delta function with. In particular, I have not made any incorrect statements in my answer. – pre-kidney Jun 28 '19 at 23:35
  • I would say that the first formula in your answer is wrong (in the sense that no one uses that definition, change of variables breaks down) and the claim that there is no definition with the desired properties is wrong (we need to make certain assumptions on the function, but the same can be said about, for example, the definition of the derivative). – Maxim Jun 29 '19 at 13:28
  • The definition of the derivative includes the necessary assumption, which is that the limit in question exists. I cannot recall seeing any definition of a derivative that did not state this assumption. In this case, it is less trivial to come up with general conditions and for this reason it is even more important to include them... – pre-kidney Jun 29 '19 at 22:43
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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.

orion
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  • 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. – pre-kidney Jun 28 '19 at 18:41
  • 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. – pre-kidney Jun 28 '19 at 18:46
  • @pre-kidney $\delta(t) = \lim_{n \to \infty} n \ 1_{t \in [0,1/n]}$ yields $$\iint_\Bbb{R^2} \delta(x^2+y^2-R^2)dxdy =\lim_{n \to \infty} n \iint_\Bbb{R^2} 1_{x^2+y^2 -R^2\in [0,0+1/n]} dxdy= \lim_{n \to \infty} n ( \pi (R^2+1/n)-\pi R^2) = \pi$$ – reuns Jun 29 '19 at 00:46
  • @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... – pre-kidney Jun 29 '19 at 04:05
  • @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 $C^0(\Bbb{R})$ – reuns Jun 29 '19 at 15:02
  • @reuns since this is math.SE and not physics.SE, a definition should have all assumptions explicitly stated or it is not a mathematical definition at all!! – pre-kidney Jun 29 '19 at 22:39
  • @pre-kidney It is exactly what I did : I gave a very simple limit definition of "expressions involving $\delta$". Please check it again and compare with the alternatives – reuns Jun 29 '19 at 22:42
  • If you are referring to your comment above - in fact, what I am pointing out is that if you take your first equation as the definition of $\delta(t)$, it does not follow that the second equation holds. And indeed, since $\delta(t)$ is not a function, there is no definition of $\delta(x^2+y^2-R^2)$ unless you give one... this is basic stuff – pre-kidney Jun 29 '19 at 22:45
  • @pre-kidney It seems you don't realize this is what I did : I gave definitions. Obviously I'll use the same definition for $x^2+y^2-R^2$ or $e^x-4\cos(y)$ – reuns Jun 29 '19 at 22:50
  • Let us continue this discussion in chat. – pre-kidney Jun 29 '19 at 22:50
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I would just rely on the standard formula for integration with a delta function: \begin{equation} \int_{x_1}^{x_2} f(x) \delta(x-a) \, dx = \begin{cases} f(a), & \text{if $x_1 < a < x_2$}, \\ 0, & \text{otherwise}. \end{cases} \end{equation} You can also consider the integration range in the left-hand side as a path $\mathcal{C}$ on the complex plane, and then it gives a non-zero result only when the path $\mathcal{C}$ passes through the point $a$.

I start from the OP's result: \begin{equation} \begin{split} I &\equiv \int_{-\infty}^\infty \int_{-\infty}^\infty \delta(x^2+y^2-R^2) \, dx dy \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{1}{2\sqrt{R^2-y^2}} \Biggl[ \delta\left(x-\sqrt{R^2-y^2}\,\right) + \delta\left(x+\sqrt{R^2-y^2}\,\right) \Biggr] dx dy. \end{split} \end{equation} You can integrate it over $x$ by the delta function for each term, but it leads to a non-zero result only when $\sqrt{R^2-y^2}$ is real; otherwise, the integration path of $x$ does not pass through $\sqrt{R^2-y^2}$ (or -$\sqrt{R^2-y^2}$ for the second term). For this reason, you will get a constraint $-R<y<R$. The two terms give the same result, so the factor $2$ in the denominator cancels. The final result is: \begin{equation} I = \int_{-R}^R \frac{dy}{\sqrt{R^2-y^2}} = \pi. \end{equation}

tueda
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