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Problem. Evaluate $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy$$

My Solution. A hint is given that "Use $\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$ "

Now,

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-5(x-\frac{3y}{5})^2}e^{-\frac{16}{5}y^2}dxdy$

But here the range of integration is whole $\mathbb{R^2}$. I know the definition of double integral on a plane bounded region in $\mathbb{R^2}$. So How should I calculate this last integral?

Can I proceed by repeatedly integrate w.r.to $x$ and $y$ like the following:

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-5(x-\frac{3y}{5})^2}e^{-\frac{16}{5}y^2}dxdy=\int_{-\infty}^{\infty}e^{-\frac{16}{5}y^2}(\int_{-\infty}^{\infty}e^{-5(x-\frac{3y}{5})^2}dx)dy=\int_{-\infty}^{\infty}e^{-\frac{16}{5}y^2}(\sqrt{\frac{\pi}{5}})dy=\sqrt{\frac{\pi}{5}}\sqrt{\frac{\pi}{16/5}}=\frac{\pi}{4}$

Is this right approach? If so how?

sigma
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  • One usually transforms $(x,y)$ to some $(u,v)$ to diagonalise the quadratic form in the exponent. Several similar questions have been asked here before including the general one. – StubbornAtom Aug 05 '18 at 11:25
  • No, you can not lose the variable $y$ upon integrating with respect to $x$, if the integral converges. – ty. Aug 05 '18 at 11:26
  • Then How should I proceed..? – sigma Aug 05 '18 at 11:27
  • Similar questions: https://math.stackexchange.com/questions/877711/evaluate-int-infty-infty-int-infty-inftye-frac12x2-xyy, https://math.stackexchange.com/questions/384732/find-the-value-of-int-infty-infty-int-infty-inftye-x2xyy2. – StubbornAtom Aug 05 '18 at 11:33
  • @StubbornAtom..in that post Can you please tell me how Matthias use the transformation $x+1/2y=x'$ while the other variable $y$ is remain unchanged....$y$ should be also changed according the transformation $x+1/2y=x'$ ..!! – sigma Aug 05 '18 at 11:49
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    @IndrajitGhosh He has used the change of variables $(x,y)\to(x',y')$ such that $x'=x-\frac{1}{2}y$ and $y'=y$, which is the same as saying $(x,y)\to(x',y)$ where $x'=x-\frac{1}{2}y$. – StubbornAtom Aug 05 '18 at 12:00
  • oh... Thank you so much.. – sigma Aug 05 '18 at 12:05
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    An explicit diagonalization is not needed, see https://math.stackexchange.com/questions/2629686/calculate-int-int-e-e5x22xyy2da/2630075#2630075 where $$\iint_{\mathbb{R}^2}e^{-Q(x,y)},dx,dy = \frac{\pi}{\sqrt{\det M_Q}}$$ is proved only assuming that the matrix $M_Q$ (associated to the quadratic form $Q$) is positive definite. It gives that the outcome here is just $\frac{\pi}{\sqrt{5^3-3^2}}=\frac{\pi}{4}$. – Jack D'Aurizio Aug 05 '18 at 16:26

3 Answers3

2

$$ 5x^2-6xy-5y^2 \equiv 8X^2+2Y^2 $$

This is obtained making a change of variables (rotation) to eliminate the cross product $x y$

now

$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(8X^2+2Y^2)}dXdY $$

and then

$$ \int_{-\infty}^{\infty}e^{-8X^2}dX = \frac{\sqrt{\pi}}{2\sqrt 2}\\ \int_{-\infty}^{\infty}e^{-2Y^2}dY = \sqrt{\frac{\pi }{2}}\\ $$

hence

$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy =\frac{\pi}{4} $$

Cesareo
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Hint: With polar substitution $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-5r^2+3r^2\sin2\theta}\ r\ drd\theta$$ Can you proceed?

Nosrati
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$$I=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-5x^2}e^{6xy}e^{-5y^2}dxdy=\left(\int_{-\infty}^{\infty}e^{-5x^2}dx\right)\left(\int_{-\infty}^{\infty}e^{-5y^2}dy\right)\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{6xy}dxdy\right)=\left(\int_{-\infty}^{\infty}e^{-5x^2}dx\right)^2 \left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{6xy}dxdy\right)$$ I think this is true, and the first integral left is easy so it would just mean solving the second. Correct me if I'm wrong.

Henry Lee
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