#### hgforum

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

$$\displaystyle \int_{0}^{b}\int_{0}^{b}\ldots\int_{0}^{b} \mbox{exp}\left(c(x_1+x_2+\ldots+x_N)^2\right)dx_1 dx_2\ldots dx_N, c>0$$
Any help would be appreciated. Thanks.

#### HallsofIvy

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

#### hgforum

Can't we use the following technique for N=1:

Set $$\displaystyle I=\int_{0}^b e^{x^2}dx$$. Then $$\displaystyle I^2=\int_{0}^b \int_{0}^b e^{x^2+y^2}dx dy$$. Now we can at least say that $$\displaystyle \int_{0}^{2\pi} \int_{0}^{b/2} e^{r^2}r dr d\theta\leq I^2\leq \int_{0}^{2\pi} \int_{0}^b e^{r^2}r dr d\theta$$. 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

MHF Helper
Can't we use the following technique for N=1:

Set $$\displaystyle I=\int_{0}^b e^{x^2}dx$$. Then $$\displaystyle I^2=\int_{0}^b \int_{0}^b e^{x^2+y^2}dx dy$$. Now we can at least say that $$\displaystyle \int_{0}^{2\pi} \int_{0}^{b/2} e^{r^2}r dr d\theta\leq I^2\leq \int_{0}^{2\pi} \int_{0}^b e^{r^2}r dr d\theta$$.
No. This only works where we have the entire first quadrant (and then $$\displaystyle \theta$$ runs from 0 to $$\displaystyle \pi/2$$ not $$\displaystyle 2\pi$$) so that r will go to infinity for all $$\displaystyle \theta$$. 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 $$\displaystyle \theta\le \pi/4$$ r runs from 0 to $$\displaystyle \frac{b}{cos(\theta)}$$ and for $$\displaystyle \theta> \pi/4$$ r runs from 0 to $$\displaystyle \frac{b}{sin(\theta)}$$.

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.