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Math Help - Integration over R^n

  1. #1
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    Integration over R^n

    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?
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  2. #2
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    Quote Originally Posted by measureman View Post
    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 (n1)-dimensional volume) of the (n1)-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}.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    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?
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  4. #4
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    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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