I've just started learning the Dirac Delta function and its properties under integration. So I have to evaluate the following:

$$I =\int^{\,\infty}_{-\infty}dx\;\int^{\,\infty}_{-\infty}dy\; x^2\delta\left(\sqrt{x^2+y^2}-R\right)\qquad,\text{where }R>0.$$

I am aware of the property of the Delta function, as follows: $$ \int^{\,\infty}_{-\infty}dx\;\delta(x-x')f(x)=f(x')\tag{1}$$ or, more specifically, $$\int^{x'+\epsilon}_{x'-\epsilon}dx\;\delta(x-x')f(x)=f(x')\tag{2}$$

So my thought process is to first convert the expression in the delta function to a single variable, $r$, in terms of $y$ as a variable and treating $x$ as a constant. Which I will define as $y=\pm\sqrt{r^2-x^2}$. My rationale for placing a plus-minus sign is to take care of the fact that I need to integrate over negative values of $y$ as well. So I also have, as a result, $\displaystyle dy=\dfrac{\pm\, r}{\sqrt{r^2-x^2}}dr.$

My next step is substituting these into the equation (1), taking care to split the integral into the rightful domains. So I get: $$\begin{align} I =\int^{\,\infty}_{-\infty}dx\left[\int^{\infty}_{0}dr\; \dfrac{rx^2}{\sqrt{r^2-x^2}}\,\delta(r-R)+\int^{\,0}_{-\infty}dr\; \dfrac{-rx^2}{\sqrt{r^2-x^2}}\,\delta(r-R)\right] \end{align} $$

Now, I was given $R>0$, and using (2), since the integration limits are not in the range of $R$, the second integral involving the negative limits should be $0$. I am left with the first integral, which, upon using (1), I obtain:

$$I =\int^{\,\infty}_{-\infty}dx\, \dfrac{Rx^2}{\sqrt{R^2-x^2}}$$.

From here on, I'm stuck. Firstly, the integration does not converge to a finite value, and secondly, the current integrand, which I obtained from evaluating the integral of the delta function at $r=R$, is not the correct expression. I checked with WolframAlpha by substituting $x$ and $R$ as constants and compared it with the expression I have in the integrand. It is off by a factor of 2. Now I am beginning to wonder if my argument for ignoring the second integration (with the negative limits) is wrong, or somewhere in my steps I have a wrong concept.

I would appreciate any hints, and I will attempt to figure out the solution as we go along.

  • $\begingroup$ Where does it come from ? How do you define $f(x,y) = \delta(\sqrt{x^2+y^2}-R)$ ? $\endgroup$ – reuns Oct 30 '16 at 1:17
  • $\begingroup$ @user1952009 If you're referring to the f(x) in eqns (1) and (2), you're mistaken. In the context of the problem I'm trying to solve, $f(x)=x^2$, and $\delta(\sqrt{x^2+y^2}-R)$ is the delta function. $\endgroup$ – Troy Oct 30 '16 at 1:50
  • $\begingroup$ $\delta(x)$ is the Dirac delta, but how do you define the distribution $T(x,y)=\delta(\sqrt{x^2+y^2}-R)$ ? $\endgroup$ – reuns Oct 30 '16 at 3:13

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  1. One Approach: \begin{align} \left.\vphantom{\LARGE a}I\,\right\vert_{\ R\ >\ 0} & = \int_{-\infty}^{\infty}\dd x \int_{-\infty}^{\infty}\dd y\,\, x^{2}\,\delta\pars{\root{x^{2} + y^{2}} - R} \\[5mm] & = \int_{0}^{2\pi}\int_{0}^{\infty}\rho^{2}\cos^{2}\pars{\theta}\, \delta\pars{\rho - R}\,\rho\,\dd\rho\,\dd\theta \\[5mm] & = \underbrace{\bracks{\int_{0}^{\infty}\rho^{3} \delta\pars{\rho - R}\,\dd\rho}}_{\ds{R^{3}}}\ \underbrace{% \bracks{\int_{0}^{2\pi}\cos^{2}\pars{\theta}\,\,\dd\theta}}_{\ds{\pi}} =\ \bbox[10px,#ffe,border:1px dashed navy]{\ds{\pi R^{3}}} \end{align}
  2. Another Approach:
    Due to the $\delta$ presence; the integral is evaluated, indeed, over $\ds{\pars{-R,R}^{2}}$ and, in addition,

    $\ds{\verts{\root{R^{2} - x^{2}}} < R}$. Namely, \begin{align} \left.\vphantom{\LARGE a}I\,\right\vert_{\ R\ >\ 0} & = \int_{-\infty}^{\infty}\dd x \int_{-\infty}^{\infty}\dd y\,\, x^{2}\,\delta\pars{\root{x^{2} + y^{2}} - R} \\[5mm] & = \int_{-R}^{R}x^{2}\int_{-R}^{R}\bracks{% {\delta\pars{y + \root{R^{2} - x^{2}}} \over \verts{y/\root{x^{2} + y^{2}}}} + {\delta\pars{y - \root{R^{2} - x^{2}}} \over \verts{y/\root{x^{2} + y^{2}}}}} \,\dd y\,\dd x \\[5mm] & = \int_{-R}^{R}x^{2} \pars{{R \over \root{R^{2} - x^{2}}} + {R \over \root{R^{2} - x^{2}}}}\,\dd x = 4R\ \overbrace{\int_{0}^{R}{x^{2} \over \root{R^{2} - x^{2}}}\,\dd x} ^{\ds{{1 \over 4}\,\pi R^{2}}} \\[5mm] & =\ \bbox[10px,#ffe,border:1px dashed navy]{\ds{\pi R^{3}}} \end{align}
  • $\begingroup$ you didn't define $\delta(\sqrt{x^2+y^2}-R)$, whereas this is the OP's question $\endgroup$ – reuns Oct 30 '16 at 1:17
  • $\begingroup$ Felix, thanks for the reply. I somewhat get your first approach, not so much the second approach (how you modified the argument in the delta function). Nevertheless, I am interested in understanding where I went wrong with my solution. I feel like it is similar to the way you went about the first approach. It would be great if you had any insight on this. Thanks! $\endgroup$ – Troy Oct 30 '16 at 2:15
  • $\begingroup$ @TroyG Approach-1 is cylindrical coord.: $x = \rho\cos\left(\theta\right)\ \mbox{and}\ y = \rho\sin\left(\theta\right)$. It seems to be the 'natural' choice. Approach-2 used the identity: $$ \int_{-\infty}^{\infty}\delta\left(\mathrm{f}\left(x\right)\right)\,\mathrm{d}x = \sum_{n}{\mathrm{f}\left(x_{n}\right) \over \left\vert\mathrm{f}'\left(x_{n}\right)\right\vert} $$ $\left\{x_{n}\right\}$ are the roots of $\mathrm{f}\left(x\right) = 0$. For instance, $\sqrt{x^{2} + y^{2}} - R = 0$ has, as a function of $y$, $\color{#f00}{2}\ roots$. Namely, $\color{#f00}{\pm}\sqrt{R^{2} - x^{2}}$ $\endgroup$ – Felix Marin Oct 30 '16 at 2:37
  • $\begingroup$ @FelixMarin I see. Any idea why my substitution for $y$ is not appropriate? That I got an infinite, non-converging integral while approaching the problem with polar coordinates yielded a finite answer? $\endgroup$ – Troy Oct 30 '16 at 2:45

Using the fact that

\begin{align} \int_{\mathbb{R}^n} f(x)\delta(g(x))\ dx = \int_{g^{-1}(0)} \frac{f(x)}{|\nabla g|}\ d\sigma(x) \end{align}

we see that \begin{align} \int_{\mathbb{R}^2} x^2\delta\left(\sqrt{x^2+y^2}-R\right)\ dA = \int_{S_R} \frac{x^2}{|(x, y)/\sqrt{x^2+y^2}|}\ d\sigma = \int_{S_R} x^2\ d\sigma. \end{align} Using polar coordinates, we have that \begin{align} \int_{S_R} x^2\ d\sigma = R^3 \int^{2\pi}_0 \cos^2\theta\ d\theta= \pi R^3 . \end{align}

  • $\begingroup$ Hi, I have not seen the first equation you quoted before. What exactly does that mean? What is $\sigma$? And is that grad(g)? $\endgroup$ – Troy Oct 29 '16 at 9:50
  • $\begingroup$ @TroyG Now I understand what you meant when you ask "is that grad(g)?". I was a little careless. $\endgroup$ – Jacky Chong Oct 30 '16 at 3:06

Given that $R>0$, we are asked to evaluate the integral $$I=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}x^{2}\delta\left(\sqrt{x^{2}+y^{2}}-R\right)dxdy$$ The transformation into polar coordinates is given by $$x = r\cos\varphi\\ y = r\sin\varphi$$ where the Jacobian is $$J=\left|\begin{array}{cc} \cos\varphi & -r\sin\varphi\\ \sin\varphi & r\cos\varphi \end{array}\right|=r\left(\cos^{2}\varphi+\sin^{2}\varphi\right)=r$$ Therefore, the differentials are related by $$dxdy=r~drd\varphi$$ After substitution of $x$ and $y$, together with the corresponding differentials, we obtain the integral in polar form \begin{eqnarray} I &=& \int\limits _{0}^{2\pi}\int\limits _{0}^{\infty}r^{2}\cos^{2}\left(\varphi\right)\delta\left(r-R\right)\,r\,drd\varphi \\ &=& \int\limits _{0}^{2\pi}\int\limits _{0}^{\infty}r^{3}\cos^{2}\left(\varphi\right)\delta\left(r-R\right)\,drd\varphi \\ &=& \int\limits _{0}^{2\pi}R^{3}\cos^{2}\left(\varphi\right)\,d\varphi \\ &=& R^{3}\left[\frac{1}{2}\varphi+\frac{1}{4}\sin\left(2\varphi\right)\right]\Bigg\vert_{0}^{2\pi} \\ &=& \pi R^{3} \end{eqnarray}


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