Double integral over $\mathbb{R}^2$

Question

I'm trying to compute $$\iint_{\mathbb{R}^2} (x^2+y^2+1)^{\frac{-3}{2}} \, dx dy$$

I know that it's a sensible idea to use polar coordinates here and so I want to look at $$\iint_{\mathbb{R}^2} (r^2+1)^{\frac{-3}{2}} r \, dr d\theta$$

What would be my upper limit for $r$ though? Would it be $\infty?$

Since you are integrating over the whole $\mathbb{R}^2$, I guess the upper limit should indeed be $\infty$. — ddsLeonardo, May 20 '16 at 14:05
Yes, the limit would be $\infty$ since you are integrating over the entire plane. — John Wayland Bales, May 20 '16 at 14:05
Notice, however, that your function is undefined on and inside the unit circle. — John Wayland Bales, May 20 '16 at 14:08
@sksks52 What is the problem with the calculation of the integral ? $-\left(r^2+1\right)^{-1/2}$ is an anti-derivative of $r\left(r^2+1\right)^{-3/2}$, so one can formally integrates. — Nicolas, May 20 '16 at 15:55
@Nicolas I have done that now. Should the answer be $2 \pi$? — , May 20 '16 at 15:57

score 1 · Accepted Answer · answered May 20 '16 at 16:28

1

$$\iint_{\mathbb{R}^2}(x^2+y^2+1)^{-3/2}\,dx\,dy = 2\pi\int_{0}^{+\infty}\frac{\rho\,d\rho}{(\rho^2+1)^{3/2}} = \pi\int_{0}^{+\infty}\frac{du}{(1+u)^{3/2}}=\color{red}{2\pi}. $$

answered May 20 '16 at 16:28

Jack D'Aurizio

361,689

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

With the coordinate transformation $(r,\phi) \rightarrow (r \, \sin \phi, r \, \cos \phi)$ you map $[0,\infty) \times [0,2\pi)$ to $\mathbb{R} \times \mathbb{R}$ so the boundaries in the second formula would be

$$\int_0^{2\pi} d\phi \int_0^\infty dr \ldots$$

As mentioned the the comment your function is not bounded on the unit circle (If a function $f(x)$ is Riemann integrable on $[a,b]$, is $f(x)$ bounded on $[a,b]$?).

Double integral over $\mathbb{R}^2$

2 Answers2