# Integration over regular simplex with Gaussian Function

Let $T_n$ be the regular n-dimensional simplex centered at the origin. Please see the diefinition in http://en.wikipedia.org/wiki/Simplex#Cartesian_coordinates_for_regular_n-dimensional_simplex_in_Rn I want to integrate over $T_n$ with Gaussian function, i.e., \begin{align} \int_{T_n} \exp \left( -\frac{x_1^2 + x_2^2 + \cdots + x_n^2}{2} \right) dx_1 dx_2 \cdots dx_n. \end{align} For example, if $n=2$, the regular simplex becomes a equilateral triangle centered at origin. I tried the above equation for $n=2$. But I failed to obtain the explicit value.

What I tried is to divide two region into the distance from the origin is less(more) than $\sqrt{3}/6$. The first region does not hit the boundaries of the triangle, so the integral can be written as,

\begin{align} 2 \pi \int_0^{\frac{\sqrt{3}}{6}} r \exp \left(-\frac{r^2}{2}\right) \, dr = 2 \left(1-\frac{1}{\sqrt[24]{e}}\right) \pi \end{align}

The second region, instead of considering all angle $2\pi$, only $6 \sin^{-1}(1/2\sqrt{3}r) - \pi$ radians should be covered. Therefore,

\begin{align} \int_{\frac{1}{2 \sqrt{3}}}^{\frac{1}{\sqrt{3}}} e^{-\frac{r^2}{2}} r \left( 6\sin ^{-1}\left(\frac{1}{2 \sqrt{3} r}\right) - \pi \right)\, dr \end{align} should be evaluated. But I failed to perform the above integration.

Furthermore, I want to generalize it to the arbitrary dimension $n$. This integration will be important in my research. Let me know anything related to the problem. Thanks.

You might check the proof of

Cartesian coordinates for vertices of a regular 16-simplex?

for a formula for the regular simplex vertices. Then you can move the midpoint to the origin.

But how do you plan to rotate the simplex at the origin?

I interpret the integral as follows:

$$I(\lambda) = n! \int_0^{x_2} dx_1 \int_0^{x_3} dx_2 \cdots \int_0^{x_n} dx_{n-1} \int_0^1 dx_n \exp{\left (-\frac{\lambda}{2}\sum_{j=1}^n x_j^2 \right )}$$

I also added a $\lambda$ for generalization. Supposedly we are interested in $I(1)$. The $n!$ term has been introduced such that $I(0)=1$.

One then has

$$I(\lambda) = \left [ \sqrt{\frac{\pi}{2 \lambda}} \mathrm{erf} \left ( \sqrt{\frac{\lambda}{2}} \right ) \right ]^n \ \ .$$