Finding when a double integral is convergent Given is the integral 
$$\iint_{\mathbb{R}^2} \frac{1}{(1+x^2+y^2)^k}\,dx \, dy$$
the question asks for the values of $k$ for which the integral will converge, and in turn find the value which the integral converges to. Using $k=1$ shows that it diverges, but I'm not sure how I should go about finding the values for which it converges. Thanks in advance for any help.
 A: Hint. By changing from cartesian coordinates  to polar ones, we obtain
$$ \iint_{\mathbb{R}^2} \frac{1}{(1+x^2+y^2)^k}\,dx\, dy = \int_{\theta=0}^{2\pi} \int_{r=0}^\infty \frac{1}{(1+r^2)^k}\,r \, dr \, d\theta = 2\pi\int_0^\infty \frac{r}{(1+r^2)^k} \, dr.$$
Now, as $r\to +\infty$,
$$\frac r {(1+r^2)^k}\sim \frac{1}{r^{2k-1}}.$$
So for which $k$ the integral is convergent? 
For such $k$, you should be are able to evaluate the integral by noting that
$$\int \frac{r}{(1+r^2)^k} \, dr=\frac{1}{2}\int (1+r^2)^{-k} d(1+r^2) = \frac{(1+r^2)^{-k+1}}{2(-k+1)}+C.$$
A: As an addendum to Robert Z' answer, 
$$ 2\pi\int_{0}^{+\infty}\frac{r\,dr}{(1+r^2)^k}\stackrel{r\mapsto\tan\theta}{=} 2\pi\int_{0}^{\pi/2}(\sin\theta)\left(\cos\theta\right)^{2k-3}\,d\theta$$
is clearly convergent as soon as $k>1$ and it is a decreasing and log-convex function of the $k$ variable.
It equals $\frac{\pi}{k-1}$.
A: \begin{align}
& \iint\limits_{\mathbb{R}^2} \frac 1 {(1+x^2+y^2)^k}\,dx\, dy = \int_0^{2\pi} \int_0^\infty \frac{1}{(1+r^2)^k}\,r \, dr \, d\theta \\[10pt]
= {} & 2\pi\int_0^\infty \frac{r}{(1+r^2)^k} \, dr = \pi \int_1^\infty \frac{du}{u^k} \qquad (\text{where } u = 1+r^2, \text{ so } du = 2r\,dr).
\end{align}
