# How to find this integral…

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\left(3x^2+2\sqrt2xy+3y^2\right)}dxdy$$

I have no idea how to integrate this function. If the middle $xy$ term would not have been present it would have been easy. But the $xy$ term is causing a problem.

• Law of exponents? – John Smith Feb 8 '17 at 19:12
• I am using that but the xy term is making problem in integration to me – Abhishek Chandra Feb 8 '17 at 19:16
• Did you try switching to polar coordinates? – Umberto P. Feb 8 '17 at 19:21
• Hint: change variables to (u, v) such that the exponent is of the form $-(au^2 + bv^2)$, for some positive constants $u$ and $v$. – Michael Lugo Feb 8 '17 at 19:21
• – Jack D'Aurizio Feb 8 '17 at 20:22

$$I=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(3x^2+2\sqrt2xy+3y^2)}dxdy$$
Let $x=r\cos\theta$, $y=r\sin\theta$. Converting to polar form, \begin{align}I&=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-(3r^2\cos^2\theta+2\sqrt2r^2\sin\theta\cos\theta+3r^2\sin^2\theta)}rdrd\theta\\ &=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2(3+2\sqrt2\sin\theta\cos\theta)}rdrd\theta\\ &=\frac12\int_{0}^{2\pi}\int_{0}^{\infty}e^{-u(3+2\sqrt2\sin\theta\cos\theta)}dud\theta\\ &=\frac12\int_{0}^{2\pi}-\frac1{(3+2\sqrt2\sin\theta\cos\theta)}\left[e^{-u(3+2\sqrt2\sin\theta\cos\theta)}\right]^{\infty}_0d\theta\\ &=\frac12\int_0^{2\pi}\frac1{3+2\sqrt2\sin\theta\cos\theta}d\theta\\ &=\frac12\int_0^{2\pi}\frac1{3+\sqrt2\sin2\theta}d\theta\\ &=\frac14\int_0^{4\pi}\frac1{3+\sqrt2\sin v}dv\\ &=\frac12\int_0^{2\pi}\frac1{3+\sqrt2\sin v}dv\\ &=\frac12\frac{2\pi}{\sqrt{3^2-2}}\\ &=\frac{\pi}{\sqrt7}\end{align}
• I believe in your first line when switching to polar coordinates, the $y$ term should become $r^2\sin^2\theta$, although it looks like the following line corrected itself. – Brian J Feb 8 '17 at 21:28
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{x \equiv u + v}$ and $\ds{y = u - v}$: \begin{align} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\pars{-\bracks{3x^{2} + 2\root{2}xy + 3y^{2}}}\,\dd x\,\dd y \\[5mm] = &\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\bracks{6 + 2\root{2}}u^{2}}\exp\pars{-\bracks{6 - 2\root{2}}v^{2}}\, \ \overbrace{\verts{\partial\pars{x,y} \over \partial\pars{u,v}}}^{\ds{2}}\,\dd u\,\dd v \\[5mm] = &\ \bbx{\ds{\root{7} \over 7}\,\pi} \approx 1.1874 \end{align}
The quadratic form $3x^2+2\sqrt 2 xy + 3y^2$ can be "diagonalised" by the (orthogonal) change of variables $x=(u+v)/\sqrt 2$ and $y=(u-v)/\sqrt 2$. They give $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathrm e^{-[(3+\sqrt 2)u^2 + (3-\sqrt 2)v^2]}~\mathrm du~\mathrm dv$$
Next, take the change of coordinates $u=r\sqrt{3-\sqrt 2}\,\cos\theta$ and $v=r\sqrt{3+\sqrt 2}\,\sin\theta$. This gives $$(3+\sqrt 2)u^2 + (3-\sqrt 2)v^2 \leadsto 7r^2$$ where $-\infty < x,y < \infty$ becomes $0 < r< \infty$ and $0<\theta <2\pi$. Moreover, $\mathrm du \wedge \mathrm dv = r\sqrt 7~\mathrm dr \wedge \mathrm d\theta.$ Hence \begin{eqnarray*} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathrm e^{-[(3+\sqrt 2)u^2 + (3-\sqrt 2)v^2]}~\mathrm du~\mathrm dv &=& \sqrt 7\int_0^{2\pi}\int_0^{\infty} r\mathrm e^{-7r^2}~\mathrm dr~\mathrm d\theta \\ \\ &=& \sqrt 7\int_0^{2\pi} \left[ -\frac{1}{14}e^{7r^2}\right]_{0}^{\infty}~\mathrm d\theta \\ \\ &=& \sqrt 7\int_0^{2\pi} \frac{1}{14}~\mathrm d\theta \\ \\ &=& \frac{\sqrt 7}{7}\pi \end{eqnarray*}