Let $\sigma$ be the "normalized Lebesgue" (Haar, really...) measure on the unit sphere $S=S^{n-1} \subset \mathbb R^n$. That is, $\sigma$ has support $S$, it is uniformly distributed, and $\int_S d\sigma = 1$.
I'm having a hard time figuring out how to understand and evaluate integrals $\int_S f\left(x\right) d\sigma\left(x\right)$ of non-trivial (let's say, even and non-constant) functions $f$. This is not really "the" $\mathbb{R}^n$ Lebesgue measure or a restriction thereof, so I can't easily convert it to a Riemann integral (unless there's something I haven't noticed).
As a simple example, how do I evaluate $\int_S (x_1)^2 d\sigma(x)$ when $n=2$? Please explain in a way that I can generalize.
Note: What I'm eventually really interested in is proving that the Fourier transform of $\sigma$, that is, $\hat\sigma(y) := \int_S \exp \left(2\pi i x\cdot y \right) d\sigma(x)$, which is always real (as $\sigma$ is symmetric), is negative on some sphere (in the Fourier-space) centered at zero. This is not supposed to be hard. I'm thinking it's going to look like a "sinc" function which is negative infinitely many times when you move away from the origin, but I don't know how to actually show that. Some pointers (but not a full proof please) would be welcome :)
It sounds like you're on the wrong track to answer your question about the Fourier transform--these integrals can't be evaluated in any explicit sense (though one can derive formulas for them involving Bessel functions). The shortest solution to your problem that I can see would be to compute the integral $$\int_{{\mathbb R}^n} e^{-\beta \pi |y|^2} \hat \sigma(y) dy$$ exactly as a function of $\beta > 0$ using Fubini, then argue that the limiting behavior as $\beta \rightarrow 0^{+}$ of this integral contradicts the assumption that $\hat \sigma(y)$ is nonnegative.