Improper Multiple Integral

Question

I am trying to solve the following exam problem:

Let $s$ be a real number. Find the condition under which the improper integral $$I:=\iint_{\mathbb R^2} \frac{dxdy}{(x^2-xy+y + 1)^s}$$ converges, and obtain the values of $I$ for the values of $s$ that makes $I$ converge.

I tried to solve it in the following way.

I transform the coordinate system from $(x,y)$ to $(\xi,\eta)$, so that the quadratic form $x^2-xy+y^2$ in the denominator may become of the form $a\xi^2+b\eta^2$. Then I rewrite the integral by using a "distorted" polar coordinate (in the shape of an ellipse, not of a circle), and then $I$ is a multiple of the integral $$ J:=\int_0^\infty \frac{dr}{(r^2+1)^s}.$$

So $I$ converges if and only if $s>1/2$ (is it true?). By letting $\tan \phi := r$, I have $J = \int_0^{\pi/2}\cos^{s-2}\phi\ d\phi$.

The problem is how to evaluate the last integral for general $s>1/2$. I would appreciate if you could give a clue, or suggest some other way to attack the original problem.

Answer 1

You can get to a familiar form by performing the substitution

$$u=\frac{1}{1+r^2}$$ $$r=\left(\frac{1}{u}-1\right)^{1/2}$$ $$dr = -\frac{du}{2 u^2} \left(\frac{1}{u}-1\right)^{-1/2}$$

Then the integral is

$$\frac12 \int_0^1 \frac{du}{2 u^2} \left(\frac{1}{u}-1\right)^{-1/2} u^s = \frac12 \int_0^1 du \, (1-u)^{-1/2} u^{s-3/2}$$

The latter integral is the form of a beta function and is equal to

$$\frac{\Gamma(1/2) \Gamma(s-1/2)}{2 \Gamma(s)} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(s-1/2)}{\Gamma(s)}$$

Answer 2

Actually, you forgot the Jacobian when you performed the change of variables! Indeed, setting $$ x-\frac y2 = r\cos\theta,\qquad \frac{\sqrt 3}2 y = r\sin\theta $$ then ${\mathrm d}x\,{\mathrm d}y=\frac2{\sqrt3}\,r\,{\mathrm d}r\,\mathrm{ d}\theta$, and therefore $$ I = \frac{4\pi}{\sqrt 3}\int_0^\infty\frac{r\,\mathrm dr}{(r^2+1)^s} =\frac{2\pi}{\sqrt3(s-1)}\,,\qquad s>1, $$ as you can easily check with Mathematica.