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Math Help - Integrating an N-dim quadratic

  1. #1
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    Integrating an N-dim quadratic

    Hello,
    I have been stuck with the following integration problem for a while now:

     <br />
\int_{0}^{b}\int_{0}^{b}\ldots\int_{0}^{b} \mbox{exp}\left(c(x_1+x_2+\ldots+x_N)^2\right)dx_1 dx_2\ldots dx_N, c>0<br />
    Any help would be appreciated. Thanks.
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  2. #2
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    Since even for N= 1, that integral has no elementary formula, I doubt you are going to find anything very useful.
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  3. #3
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    Can't we use the following technique for N=1:

    Set I=\int_{0}^b e^{x^2}dx. Then I^2=\int_{0}^b \int_{0}^b e^{x^2+y^2}dx dy. Now we can at least say that \int_{0}^{2\pi} \int_{0}^{b/2} e^{r^2}r dr d\theta\leq I^2\leq \int_{0}^{2\pi} \int_{0}^b e^{r^2}r dr d\theta. Furthermore, the upper and lower bounds can be computed explicitly.

    Isn't this correct firstly? and if so, can't we extend this to more dimensions?

    Thanks.
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  4. #4
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    Quote Originally Posted by hgforum View Post
    Can't we use the following technique for N=1:

    Set I=\int_{0}^b e^{x^2}dx. Then I^2=\int_{0}^b \int_{0}^b e^{x^2+y^2}dx dy. Now we can at least say that \int_{0}^{2\pi} \int_{0}^{b/2} e^{r^2}r dr d\theta\leq I^2\leq \int_{0}^{2\pi} \int_{0}^b e^{r^2}r dr d\theta.
    No. This only works where we have the entire first quadrant (and then \theta runs from 0 to \pi/2 not 2\pi) so that r will go to infinity for all \theta. Integrating from 0 to b gives a rectangle in the xy-plane and, in polar coordinates, r will run from 0 to either the right or top edge of the rectangle and that r will depend on theta. For \theta\le \pi/4 r runs from 0 to \frac{b}{cos(\theta)} and for \theta> \pi/4 r runs from 0 to \frac{b}{sin(\theta)}.

    Furthermore, the upper and lower bounds can be computed explicitly.

    Isn't this correct firstly? and if so, can't we extend this to more dimensions?

    Thanks.
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