Compute Integral over Implicit Domain

Question

How would I proceed in computing:

$$\int_D e^{-(x_1^2 + ... + x_n^2)/2} dx \qquad \qquad \text{ with } \quad D=\left\{x \in \mathbb R^n: \left|\sum_{i=1}^n x_i\right| \le \sqrt{n}\right\}.$$

I need to somehow get rid of the implicitly given domain. Perhaps, by using properties of the Gaussian?

Answer 1

This is best understood in probabilistic terms.

Since the pdf of the standard normal distribution is given by $$ f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2} \tag{1}$$ we have that $$ P = \frac{1}{(2\pi)^{n/2}}\int_{D}e^{-(x_1^2+\ldots+x_n^2)/2}\,d\mu \tag{2}$$ is the probability that the sum of $n$ independent standard normal variables lies in $[-\sqrt{n},\sqrt{n}]$.

This sum is just a normal variable with mean zero and $\sigma^2=n$, hence:

$$ P = \frac{1}{\sqrt{2n\pi}}\int_{-\sqrt{n}}^{\sqrt{n}}\exp\left(-\frac{x^2}{2n}\right)\,dx =\frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{-y^2/2}\,dy=\operatorname{Erf}\left(\frac{1}{\sqrt{2}}\right)\tag{3}$$ and:

$$ \int_{D}e^{-(x_1^2+\ldots+x_n^2)/2}\,d\mu = (2\pi)^{n/2}\operatorname{Erf}\left(\frac{1}{\sqrt{2}}\right).$$