Evaluate $\int^{\infty}_{-\infty} \int^{\infty}_{-\infty} e^{-\left(3x^2+2 \sqrt 2 xy+3y^2\right)} \,dx\,dy$

Question

Evaluate $$\displaystyle\int^{\infty}_{-\infty} \int^{\infty}_{-\infty} e^{-\left(3x^2+2 \sqrt 2 xy+3y^2\right)} \,\mathrm dx\,\mathrm dy\,$$

My attempt :$ 3x^2+2 \sqrt 2xy+3y^2=3\left (x+\dfrac{\sqrt 2}{3}y\right )^2+y^2.$

$u=\sqrt{3}\left (x+\dfrac{\sqrt 2}{3}\right )y$ and $v=y$. hen

$$ \mathrm du\,\mathrm dv = \left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,\mathrm dx\,\mathrm dy = \left|\begin{matrix} \sqrt 3 & 0 \\ \frac{\sqrt 3}{2}& 1 \end{matrix}\right|\,\mathrm dx\,\mathrm dy = \sqrt 3 \,\mathrm dx\,\mathrm dy $$

And

$$ I = \dfrac{1}{ \sqrt3 }\iint_{\Bbb R^2} e^{-(u^2+v^2)}\, \mathrm du\,\mathrm dv $$

Switch to polar coordinates (see solution here)

Now my final answer is $\dfrac{\pi}{\sqrt 3}$

Is it correct ?

Answer 1

The approach is right, but there are some minor mistakes in calculations.

$$ 3x^2+2\sqrt2xy+3y^2=3\left (x+\frac{\sqrt2}3y\right )^2+\frac73y^2. $$ So we take $u=\sqrt3\left (x+\dfrac{\sqrt2}3y\right )$ and $v=\dfrac{\sqrt7}{\sqrt3}y$. Then $$ \left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,\mathrm dx\,\mathrm dy = \left|\begin{matrix} \sqrt 3 & \dfrac{\sqrt 2}{\sqrt3} \\ 0& \dfrac{\sqrt7}{\sqrt3} \end{matrix}\right|\,\mathrm dx\,\mathrm dy. $$

Thus the final answer should be $\dfrac\pi{\sqrt7}$.

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Hope this helps.