TITLE: Converting double integrals to polar form: what would the limits be here?
QUESTION [9 upvotes]: So I have the following integral:
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} {{{x^2}dxdy}\over (1 + \sqrt{x^2 + y^2})^5}$$
I know that converting the integral using polar coordinates gives:
$$\int \int {{r^2 \cos^2 \theta} \over {(1 + r)^5}} rdrd \theta$$
I'm assuming $r$ is going from $0$ to infinity.
But what about $\theta$?

REPLY [11 votes]: Note that in order to cover $\mathbb{R}^2$, $r$ extends from $0$ to $\infty$ and $\theta$ spans an entire period of $\sin(\theta)$ and $\cos(\theta)$.  (For example, the first quadrant alone is covered by $\theta \in [0,\pi/2]$, $r\in [0,\infty)$.) 
Hence, 
$$\begin{align}
\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{x^2}{(1+\sqrt{x^2+y^2})^5}\,dx\,dy&=\int_0^{2\pi}\int_0^\infty \frac{r^2\cos^2(\theta)}{(1+r)^5}r\,dr\,d\theta\\\\
&=\underbrace{\left(\int_0^{2\pi}\cos^2(\theta)\,d\theta\right)}_{=\pi}\underbrace{\left(\int_0^\infty \frac{r^3}{(1+r)^5}\,dr\right)}_{=1/4}\\\\
&=\frac{\pi}{4}
\end{align}$$