Q: Convergence of improper double integral Consider the improper double integral  
$$I=\iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}{(x^2+4y^2)^{\alpha}}\ dx\ dy$$
I want to determine the range of $\alpha$ , such that the integral above is convergent. Using transformation to polar coordinates I get $ 
I=\iint_{r>0}\frac{rsin(r^2)}{r^{2\alpha}(1+3sin^2\theta)^{\alpha}}\ dr\ d\theta
$ (I am not sure, if $r$ - determinant of the Jacobian is applied here). So it is $ 
I=\iint_{r>0}\frac{sin(r^2)}{r^{2\alpha-1}(1+3sin^2\theta)^{\alpha}}\ dr\ d\theta
$. For $\lim{r\to 0} \ \alpha>2 \ (2\alpha-3<1) $ and for $\lim{r\to \infty} \ \alpha<1 \ (2\alpha-1>1)$. But I don't know what to do with $\frac{1}{(1+3sin^2\theta)^{\alpha}}$ and how to 
substantiate this range of $\alpha$. Any help would be appreciated.
Edit: $$\frac{1}{4^{\alpha}}{\iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}{(x^2+y^2)^{\alpha}}\ dx\ dy} \leq \iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}
 {(x^2+4y^2)^{\alpha}}\ dx\ dy \leq {\iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}{(x^2+y^2)^{\alpha}}\ dx\ dy}$$ Can I do it like that or should I add module?
 A: It is more practical to "confine the mess to the numerator" then switch to polar coordinates:
$$\begin{eqnarray*} I &=& \frac{1}{2}\iint_{\mathbb{R}^2\setminus\{(0,0)\}}\frac{\sin\left(x^2+\frac{1}{4}y^2\right)}{(x^2+y^2)^{\alpha}}\,dx\,dy \\[0.2cm]&=&\frac{1}{2}\,\text{Im}\int_{0}^{+\infty}\int_{0}^{2\pi}\frac{e^{i\rho^2\left(\cos^2\theta+\frac{1}{4}\sin^2\theta\right)}}{\rho^{2\alpha-1}}\,d\theta\,d\rho\\[0.2cm]&=&\frac{1}{4}\,\text{Im}\int_{0}^{+\infty}\int_{0}^{2\pi}\frac{e^{iz\left(\cos^2\theta+\frac{1}{4}\sin^2\theta\right)}}{z^{\alpha-1}}\,d\theta\,dz\tag{1}\end{eqnarray*}$$
and $\cos^2\theta+\frac{1}{4}\sin^2\theta$ is bounded between $\frac{1}{4}$ and $1$ for every $\theta$. By Dirichlet's test the integral $\int_{0}^{+\infty}\frac{e^{ikz}}{z^{\alpha-1}}\,dz$ is convergent for any $\alpha\in(1,2)$, and with such assumption:
$$ I = \frac{\Gamma(2-\alpha)}{4}\,\sin\left(\frac{\pi \alpha}{2}\right)\int_{0}^{2\pi}\frac{d\theta}{\left(\cos^2\theta+\frac{1}{4}\sin^2\theta\right)^{2-\alpha}} \tag{2}$$
where the last integral is finite and can be computed through the residue theorem. It is interesting to point out that the limit of the previous expression as $\alpha\to 2^-$ simply equals $\frac{\pi^2}{4}$.
