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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.

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If Maple is correct $$\int_0^\infty \frac{dr}{(r^2+1)^s} = \frac{\sqrt{\pi}}{2} \frac{\Gamma(s-1/2)}{\Gamma(s)}$$ – gammatester Aug 6 '13 at 7:59
If $s=1$, the the integral $I$ diverges up to the Maple command $$VectorCalculus:-int(1/(x^2-x*y+y^2+1), [x, y] = Rectangle(-infinity .. infinity, -infinity .. infinity))$$ $$\infty$$ – user64494 Aug 6 '13 at 8:08
Maple produces the answer $$(1/3)*Pi*(sqrt(3)*sqrt(3*RootOf(RootOf(Im(Z))^2-RootOf(Im(Z))*Z+Z^2+4)^2+16)+3‌​*RootOf(RootOf(Im(Z))^2-$$ $$RootOf(Im(Z))*Z+Z^2+4))/sqrt(3*RootOf(RootOf(Im(Z))^2-$$ $$RootOf(Im(Z))*Z+Z^2+4)^2+16)$$ in the case $s=2.$ In view of it I have doubts concerning a closed form for $I.$ – user64494 Aug 6 '13 at 8:16
I think it should be $r$ in the numerator of the integrand of $J$. In this case the right condition $s>1$ of the convergence is obtained. – user64494 Aug 6 '13 at 8:23
You can use this form of beta function to evaluate $J$ $\beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0 \![2]$. – Mhenni Benghorbal Aug 6 '13 at 20:44
up vote 3 down vote accepted

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

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

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Thank you. Under what circumstances, in general, is this substitution useful? – Pteromys Aug 6 '13 at 8:40
It is useful in converting an infinite region of integration into a finite one. – Ron Gordon Aug 6 '13 at 11:22

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