$\int_{\mathbb{R}^2} \delta(E-ax-by) x^2 dx $

Question

I am wondering how we have to integrate $\int_{\mathbb{R}^2} \delta(E-ax^2-by^2) x^2 dxdy.$ I am not familiar with this kind of delta distribution (depending on two coordinates), so I was wondering if there is a standard trick to evaluate this integral?

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\int_{{\mathbb R}^{2}}% \delta\pars{E - ax^{2} - by^{2}}x^{2}\,\dd x\,\dd y} \\[5mm]&=\int_{-\infty}^{\infty}\dd x\,x^{2}\,\Theta\pars{E - ax^{2} \over b} \times \\&\int_{-\infty}^{\infty} {\delta\pars{y + \root{\bracks{E - ax^{2}}/b}} + \delta\pars{y - \root{\bracks{E - ax^{2}}/b}} \over \verts{2by}} \,\dd y \\[5mm]&={1 \over \verts{b}} \int_{-\infty}^{\infty}\Theta\pars{E - ax^{2} \over b}x^{2}\, {\dd x \over \root{\pars{E - ax^{2}}/b}} \end{align}

There are several possibilities according to the signs of $\ds{E, a}$ and $\ds{b}$.

I'll perform an evaluation in the case $\ds{E >0\,,\ a>0\,,b>0}$: \begin{align}&\color{#66f}{\large\int_{{\mathbb R}^{2}}% \delta\pars{E - ax^{2} - by^{2}}x^{2}\,\dd x\,\dd y} ={1 \over \root{b}}\ \overbrace{% \int_{-\root{E/a}}^{\root{E/a}}{x^{2}\,\dd x \over \root{E - ax^{2}}}} ^{\ds{\color{#c00000}{x \equiv \root{E \over a}\sin\pars{\theta}}}} \\[5mm]&={2 \over \root{b}} \int_{0}^{\pi/2}{\pars{E/a}\sin^{2}\pars{\theta} \over \root{E}\cos\pars{\theta}}\, \root{E \over a}\,\cos\pars{\theta}\,\dd\theta ={2E \over a^{3/2}b^{1/2}}\ \overbrace{\int_{0}^{\pi/2}\sin^{2}\pars{\theta}\,\dd\theta}^{\ds{=\ \color{#c00000}{\pi \over 4}}} \end{align}

\begin{align}&\color{#66f}{\large\int_{{\mathbb R}^{2}}% \delta\pars{E - ax^{2} - by^{2}}x^{2}\,\dd x\,\dd y} =\color{#66f}{\large{\pi \over 2}\,{E \over a^{3/2}b^{1/2}}}\,,\qquad E>0\,,\ a>0\,,\ b>0 \end{align}

Answer 2

Use Ruffini Theorem in order to integrate first with respect to $dx$ and then with repect to $dy$, that is,

$\begin{equation*} \int_{\mathbb{R}^2} \delta(E-ax-by)x^2 dx dy = \int_{\mathbb{R}} \Big( \int_{\mathbb{R}} \delta(E-ax^2-by^2)x^2 dx \Big) dy \end{equation*}$

Now you can think the delta function as a function of $x$ only, taking $y$ constant (true by computing the first integral); use the following property of the delta function:

$\begin{equation*} \delta( f(x) ) = \sum_{x_0 \text{ zero of $f$}} \frac{\delta(x-x_0)}{|f'(x_0)|} \end{equation*}$

P.D. By using this method you must deal with square roots: in order to avoid them you can use a change of variables: try changing

$\begin{equation*} \phi : \mathbb{R}^2 \longrightarrow (0,+\infty) \times (0, 2 \pi) \end{equation*}$

given by $\phi(x,y) = (\frac{1}{\sqrt{a}}·r· \cos(t), \frac{1}{\sqrt{b}}·r· \sin(t))$, $r$ varying in $(0,+\infty)$, and $t$ in $(0, 2 \pi)$. (Don't forget the Jacobian!)