Integration over R^n

**measureman** · May 13th 2011, 02:34 AM

Hi all,

I've been trying to solve a problem which has boiled down to whether the integral

$\int_{\mathbb{R}^n} \frac{1}{(1 + |x|^2)^N} \, \mathrm{d}x$

exists for some large enough N, where $x \in \mathbb{R}^n$ .

I have been told that this does exist, but I don't know how I would prove this?

**Opalg** · May 13th 2011, 05:21 AM

Originally Posted by measureman

Hi all,

I've been trying to solve a problem which has boiled down to whether the integral

$\int_{\mathbb{R}^n} \frac{1}{(1 + |x|^2)^N} \, \mathrm{d}x$

exists for some large enough N, where $x \in \mathbb{R}^n$ .

I have been told that this does exist, but I don't know how I would prove this?

If you think in terms of computing the integral using a spherical shell method, you get

$\int_{\mathbb{R}^n} \frac{1}{(1 + |x|^2)^N} \, dx = \int_0^\infty \frac{1}{(1 + r^2)^N}S_{n-1}(r)\,dr,$

where $S_{n-1}(r)$ is the "surface area" (or more correctly the (n–1)-dimensional volume) of the (n–1)-sphere. Since $S_{n-1}(r) = k_nr^{n-1}$ for some constant $k_n$ , it follows that $\int_{\mathbb{R}^n} \frac{1}{(1 + |x|^2)^N} \, dx = k_n\int_0^\infty \frac{r^{n-1}}{(1 + r^2)^N}\,dr,$ and that converges for $N\geqslant\tfrac12(n+1)$ by comparison with the integral of $r^{-2}.$

**measureman** · May 13th 2011, 07:28 AM

Originally Posted by Opalg

If you think in terms of computing the integral using a spherical shell method, you get

$\int_{\mathbb{R}^n} \frac{1}{(1 + |x|^2)^N} \, dx = \int_0^\infty \frac{1}{(1 + r^2)^N}S_{n-1}(r)\,dr,$

where $S_{n-1}(r)$ is the "surface area" (or more correctly the (n–1)-dimensional volume) of the (n–1)-sphere. Since $S_{n-1}(r) = k_nr^{n-1}$ for some constant $k_n$ , it follows that $\int_{\mathbb{R}^n} \frac{1}{(1 + |x|^2)^N} \, dx = k_n\int_0^\infty \frac{r^{n-1}}{(1 + r^2)^N}\,dr,$ and that converges for $N\geqslant\tfrac12(n+1)$ by comparison with the integral of $r^{-2}.$

Thanks for your reply, but I know nothing about spherical shell integration (can't find much on the internet either), so I'm not too sure how this works. Is it similar to spherical polar co-ordinates?

**davismj** · May 13th 2011, 09:09 AM

Yea, think of two-dimensions. Instead of integrating over ${\mathbb_{R}}^2$ in cartesian coordinates, you integrate first over a circle of radius r, and then you integrate r from 0 to infinity.

To understand why in n-dimensions we take the (n-1) dimensional volume or surface area, notice that when you integrate the circle of radius r, you're integrating in one dimension, over a line from theta = 0 to theta = 2pi. From there you pick up the last dimension by integrating the surface area for all radii.

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