Integrating an N-dim quadratic

**hgforum** · May 22nd 2010, 06:47 AM

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.

**HallsofIvy** · May 22nd 2010, 09:22 AM

Since even for N= 1, that integral has no elementary formula, I doubt you are going to find anything very useful.

**hgforum** · May 22nd 2010, 11:24 AM

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.

**HallsofIvy** · May 23rd 2010, 02:30 AM

Originally Posted by hgforum

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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