Dirac Delta in two dimensions

Question

I would like to compute the following two dimensional integral: $$\iint f(\vec{x})\delta(|\vec{x}| -v)d^2\vec{x},$$ $$f(\vec{v})=\Bigg(\frac{m}{2\pi k_BT}\Bigg)^{3/2}\exp{\Big(\frac{-mv^2}{2k_BT}\Big)},$$ where $f$ is a smooth function. I absolutely don't know how to deal with such integrals, and I don't have any knowledge (expect the defintion) of the Dirac Delta function in higher dimensions.

Answer 1

Before doing the calculation you asked for, let me point out the Dirac delta is not a function and the symbol $ \int f(x)\delta(x)d x $ is not to be taken literally. Digging into details on this side, however, would lead to a lengthy answer. Therefore, I will stick to the physics notation, which you also use in the question.

Now lets look at your integral and start by noting, that the term $\delta(|\vec x|-v)$ is actually a one-dimensional Dirac delta. The main task therefore is, to extract its one-dimensional character. This is done, by integrating in spherical coordinates.

We work under the following two assumptions: integration is over $\mathbb R^d$ and $f:\mathbb R^d\to \mathbb R$ is continuous and rotation invariant. I will abuse notatino and write $f(|\vec r|)=f(\vec r)$ from now on. Then, integration in spherical coordinates yields \begin{align*} \int_{\mathbb R^d} f(\vec r)\delta(|\vec r|-v)d\vec r &= \int_{\mathbb R^+} dr r^{d-1} \int_{S^{d-1}}dA f(r)\delta(r-v)\\ &=v^{d-1}f(v)\int_{S^{d-1}}dA\\ &= v^{d-1}f(v)A(S^{d-1}), \end{align*} where $A(S^{d-1})$ denotes the area of the $(d-1)$-dimensional unit sphere.

Edit adding the not rotation invariant case: You can also drop the assumption, that $f$ is rotation invariant. Then, a similar calculation as above yields $$ \int_{\mathbb R^d} f(\vec r)\delta(|\vec r|-v)d\vec r = v^{d-1}\int_{S^{d-1}}f(v\vec n_A)dA, $$ where $\vec n_A$ denotes the normal vector on the unit sphere. So the volume integral reduces to a surface integral on the unit sphere by the Dirac delta.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\iint \pars{m \over 2\pi\,\mrm{k_{B}}T}^{3/2} \exp{\pars{-\,{m\verts{\vec{x}}^{2} \over 2\mrm{k_{B}}T}}} \delta\pars{\verts{\vec{x}} - v}\dd^{2}\vec{x}} \\[5mm] = &\ \pars{m \over 2\pi\,\mrm{k_{B}}T}^{3/2}\int_{0}^{2\pi}\int_{0}^{\infty} \exp{\pars{-\,{m\rho^{2} \over 2\mrm{k_{B}}T}}} \delta\pars{\rho - v}\rho\,\dd\rho\,\dd\phi \\[5mm] = &\ \bbx{\pars{m \over 2\pi\,\mrm{k_{B}}T}^{3/2} \braces{2\pi v\ \exp{\pars{-\,{mv^{2} \over 2\mrm{k_{B}}T}}}\bracks{v > 0}}} \\ & \end{align}