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Compute $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+(x-y)^2+y^2)}dxdy$.

I tried to do this by using polar coordinate. Let $x=r\cos t,\ y=r\sin t$, and then

$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+(x-y)^2+y^2)}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-2r^2}e^{2\cos t\sin t}rdrdt=\int_{0}^{2\pi}e^{\sin(2t)}\int_{0}^{\infty}e^{-2r^2}rdrdt=\frac{1}{4}\int_{0}^{2\pi}e^{\sin (2t)}dt. $$ But, I have no idea how to compute $\int_{0}^{2\pi}e^{\sin (2t)}dt$. Please give me some hint or suggestion. Thanks.

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5 Answers 5

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Expanding out $(x-y)^2$, we get $$ x^2+(x-y)^2+y^2=2(x^2-xy+y^2)$$ Then complete the square: $$ x^2-xy+y^2=\Big(x-\frac{y}{2}\Big)^2+\frac{3y^2}{4} $$ so that $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-(x^2+(x-y)^2+y^2))\;dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\Big[2\Big(x-\frac{y}{2}\Big)^2+\frac{3y^2}{2}\Big]\;dxdy$$ Using translation invariance and the Gaussian integral, this becomes $$\Big(\int_{-\infty}^{\infty}e^{-2x^2}\;dx\Big)\Big(\int_{-\infty}^{\infty}e^{-\frac{3y^2}{2}}\;dy\Big)=\sqrt{\frac{\pi}{2}}\sqrt{\frac{2\pi}{3}}=\frac{\pi}{\sqrt{3}}$$

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    $\begingroup$ You're faster than I. (+1) $\endgroup$
    – Mark Viola
    Apr 2, 2017 at 6:15
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There was an error in the OP. Notice that the term $\sin(t)\cos(t)$ was missing the multiplicative factor of $r$. Correcting this, we obtain

$$\begin{align} \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-2(x^2-xy+y2)}\,dx\,dy&=\int_0^{2\pi}\int_0^\infty e^{-2r^2(1-\sin(t)\cos(t))}\,r\,dr\,dt\\\\ &=\int_0^{2\pi} \frac{1}{4(1-\sin(t)\cos(t))}\,dt\\\\ &=\frac14\int_0^{2\pi} \frac{1}{1-\frac12\sin(2t)}\,dt\tag 1\\\\ &=\frac{\pi}{\sqrt 3}\tag 2 \end{align}$$

where in going from $(1)$ to $(2)$ we use made use of the substitution $x=\tan(t)$ to transform the integral to

$$\frac14\int_0^{2\pi} \frac{1}{1-\frac12\sin(2t)}\,dt=\frac12\int_{-\infty}^\infty\frac{1}{x^2-x+1}\,dx=\frac{\pi}{\sqrt 3}$$

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First of all ypur computation is false as some $r$ is missing before $\sin 2 \theta$. For your problem, you should diagonalize the quadratic form $x^2+(x-y)^2+y^2$ in some orthonormal basis, and the computation becomes obvious. More generaly if $q$ is a positive definite quadratic form, you can compute $\int exp -q(x) dx$ by digonalizing $q$. The general formula wil give you something like ${1\over \sqrt \delta }\int exp -q_0(x) dx$, where $q_0$ is the standard euclidian quadratic form, and $\delta$ the discriminant of $Q$, ie the volume of the ellipsoid $q\leq 1$ (in your case the area of $x^2+(x-y)^2+y^2 \leq1$).

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  • $\begingroup$ I arrived 2 minutes later than carmicahel561 $\endgroup$
    – Thomas
    Apr 2, 2017 at 6:18
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

It's better to $\underline{\mbox{take advantage}}$ of the $\bbox[5px,#efe]{integrand\ symmetries}$. Namely,

\begin{align} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\braces{x^{2} + \bracks{x - y}^{2} + y^{2}}}\,\dd x\,\dd y \\[5mm] = &\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{\bracks{x + y}^{2} + 3\bracks{x - y}^{2} \over 2}}\,\dd x\,\dd y \\[5mm] = & \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{x^{2} + 3\bracks{x - 2y}^{2} \over 2}}\,\dd x\,\dd y = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{x^{2} \over 2} - 6y^{2}}\,\dd x\,\dd y \\[5mm] = &\ \root{2}\,{1 \over \root{6}} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\pars{-x^{2} - y^{2}} \,\dd x\,\dd y = {\root{3} \over 3}\pars{\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x}^{2} = \bbx{\ds{{\root{3} \over 3}\,\pi}} \end{align}

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As shown here, if $Q$ is a positive definite symmetric matrix, $$ \int_{\mathbb{R}^n}\exp\left(- x^T Q x\right)\,d\mu = \frac{\pi^{n/2}}{\sqrt{\det Q}} \tag{A} $$ In our case we have $n=2$ and $Q=\begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}$, hence the answer is $\color{red}{\large\frac{\pi}{\sqrt{3}}}$.

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