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

$
\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
$

Any help would be appreciated. Thanks.

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

3. 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.

4. 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.