exponential integral on n-dimensional unit sphere I'd like to compute the expectation of $\exp(\sum x_i)$ under the uniform distribution on the sphere with radius $r&gt; 0$. To this end, I would like to evaluate the following integral:
$$\int_{\large{x_1, \ldots, x_n,\ \sum x_i^2\  =\  r}}\exp\left(\sum_{i=1}^n x_i\right)\,\mathrm{d}x_1 \ldots \mathrm{d}x_n $$
That is, the exponential of the sum of the (Cartesian) coordinates on the sphere.
Maple cannot do it. It is easy to find results about integrating polynomials on the sphere, and I found something about integrating just $\exp(x_1)$ on the $n$-dimensional sphere, but nothing on the exponential of the full sum.  Any help would be greatly appreciated!
If you know how to integrate $\exp(x_1)$ you know this integral, as follows. Write $e_1$ for the standard basis vector $(1, 0, 0, \dots)$ of $\mathbb{R}^n$, where we think of $x = (x_1, \dots x_n) \in \mathbb{R}^n$ as a vector, and write $v = (1, 1, 1, \dots)$ for the all-ones vector. Let me also replace your $r$ with an $r^2$ so that $r$ is the length $\| x \|$, which will make things nicer. You want to compute
$$I(v, r) = \int_{|x| = r} \exp(\langle v, x \rangle) \, dV_x$$
(where $dV_x$ is shorthand I just made up for the volume element $dx_1 \dots dx_n$) and you say you know how to compute
$$I(e_1, r) = \int_{|x| = r} \exp(\langle e_1, x \rangle) \, dV_x.$$
But by spherical symmetry it's clear that $I(v, r)$, as a function of $v$, depends only on the length of $v$! So we can in fact reduce these integrals to each other. Write $v = su$ where $\| u \| = \frac{v}{\| v \|}$ is a unit vector and $s = \| v \|$ is the length of $v$ (which in this case is $\sqrt{n}$). Then
$$\langle v, x \rangle = \langle su, x \rangle = \langle u, sx \rangle$$
and so making the change of coordinates $y = sx$ (so $y_i = sx_i$) we have that
$$I(v, r) = \int_{|y| = sr} \exp(\langle u, y \rangle) \frac{dV_y}{s^n} = \frac{1}{s^n} I(u, sr) = \frac{1}{\sqrt{n}^n} I(e_1, \sqrt{n} r).$$
If you know how to integrate $\exp(x_1)$ you know this integral, as follows. Write $e_1$ for the standard basis vector $(1, 0, 0, \dots)$ of $\mathbb{R}^n$, where we think of $x = (x_1, \dots x_n) \in \mathbb{R}^n$ as a vector, and write $v = (1, 1, 1, \dots)$ for the all-ones vector. Let me also replace your $r$ with an $r^2$ so that $r$ is the length $\| x \|$, which will make things nicer. You want to compute
$$I(v, r) = \int_{|x| = r} \exp(\langle v, x \rangle) \, dV_x$$
(where $dV_x$ is shorthand I just made up for the volume element $dx_1 \dots dx_n$) and you say you know how to compute
$$I(e_1, r) = \int_{|x| = r} \exp(\langle e_1, x \rangle) \, dV_x.$$
But by spherical symmetry it's clear that $I(v, r)$, as a function of $v$, depends only on the length of $v$! So we can in fact reduce these integrals to each other. Write $v = su$ where $\| u \| = \frac{v}{\| v \|}$ is a unit vector and $s = \| v \|$ is the length of $v$ (which in this case is $\sqrt{n}$). Then
$$\langle v, x \rangle = \langle su, x \rangle = \langle u, sx \rangle$$
and so making the change of coordinates $y = sx$ (so $y_i = sx_i$) we have that
$$I(v, r) = \int_{|y| = sr} \exp(\langle u, y \rangle) \frac{dV_y}{s^n} = \frac{1}{s^n} I(u, sr) = \frac{1}{\sqrt{n}^n} I(e_1, \sqrt{n} r).$$