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

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.

Answer

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}\,}\,}
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\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}