Finding when a double integral is convergent

Question

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.

Answer 1

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 $\frac{\pi}{k-1}$.

Answer 2

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 $$\int \frac{r}{(1+r^2)^k} \, dr=\frac{1}{2}\int (1+r^2)^{-k} d(1+r^2) = \frac{(1+r^2)^{-k+1}}{2(-k+1)}+C.$$

Answer 3

\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}