Find: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(3x^2+2\sqrt{2}xy+3y^2)}\,dxdy$$ I have no idea how to solve this,I would be thankful, if someone help me to solve this
Thank you.
Find: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(3x^2+2\sqrt{2}xy+3y^2)}\,dxdy$$ I have no idea how to solve this,I would be thankful, if someone help me to solve this
Thank you.
On
Hint: Try writing the integral as $$ \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-3\left(x+\frac{\sqrt2}{3}y\right)^2-\frac73y^2}\,\mathrm{d}x\,\mathrm{d}y\tag{1} $$ followed by a simple substitution in $x$.
Another Hint: Explain why $(1)$ is the same as $$ \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-3x^2-\frac73y^2}\,\mathrm{d}x\,\mathrm{d}y =\int_{-\infty}^\infty e^{-3x^2}\,\mathrm{d}x\int_{-\infty}^\infty e^{-\frac73y^2}\,\mathrm{d}y\tag{2} $$
On
The Idea behind it is to change the set $[3x^2-2\sqrt2xy+3y^2]$ on a circle of the form $u^2+v^2$ because we know that
$$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\,dxdy=\pi$$
In order to do that you substitute
$$x=mu-nv\space ; \space y=mu+nv$$ in the equation $3x^2-2\sqrt2xy+3y^2$. You should get at the and that
$$3x^2-2\sqrt2xy+3y^2 =(6+2\sqrt2)m^2u^2+(6-2\sqrt2)n^2v^2$$
Now you see that $m$ and $n$ must be equal to
$$m=\frac{1}{\sqrt{6+2\sqrt{2}}}\space ; \space n=\frac{1}{\sqrt{6-2\sqrt{2}}}$$
In this way if you resubstitute $ m$ and $n$ you get your desired $u^2+v^2$
To calculate your integral you have to check that the function which transformes $x$ and $y$ is a diffeomorphimus. To calculate the integral you need the Jacobian of the diffeomorphismus which is the determinant of the Jacobi Matrix which is
$$det\begin{pmatrix} \frac{1}{\sqrt{6+2\sqrt{2}}} & \frac{1}{-\sqrt{6-2\sqrt{2}}} \\\ \frac{1}{\sqrt{6+2\sqrt{2}}} & \frac{1}{\sqrt{6-2\sqrt{2}}} \end{pmatrix} = J_\phi$$
Now you integral becomes
$$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-3\left(x+\frac{\sqrt2}{3}y\right)^2-\frac73y^2}\,\mathrm{d}x\,\mathrm{d}y= J_\phi\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(u^2+v^2)}\,dudv=J_\phi\pi$$
For the last equality I've used this theorem (see under Theorem change of variables)
I hope my solution does not contain calculus error :-)
Idea: Try to modify the exponent to be a square.