Closed form for Fourier transform-like Integral on $S^{n}$

Question

Hello!

It may be a stupid question, i'm trying to find a closed form for an integral similar to a Fourier transform on $S^{n}$ but i'm stuck... Let $\alpha>0$, the integral i can't solve is

$$I(p,\alpha)=\int_{S^{n}}e^{i\alpha\left<p,q\right>}d\mu_{S^{n}}(q)$$

where $p,q\in S^{n}$, $\left< \cdot,\cdot \right>$ is the euclidean scalar product on $\mathbb{R}^{n+1}$ and $d\mu_{S^{n}}$ is the measure induced by euclidean measure on $\mathbb{R}^{n+1}$. so the question is: is there a closed form for $I(p,\alpha)$ only in terms of $p$ and $\alpha$?

Thank you in advance!

Answer 1

The integral obviously doesn't depend on $p$, since it must be rotationally invariant, and, using polar coordinates centered on $p$, one obtains $$ I(p,\alpha) = \mathrm{vol}(S^{n-1})\int_{-1}^1 e^{i\alpha t}(1{-}t^2)^{(n-2)/2}dt =\frac{2\ \pi^{n/2}}{\Gamma(n/2)}\int_{-1}^1 \cos(\alpha t)(1{-}t^2)^{(n-2)/2}dt. $$ This latter integral can be evaluated by standard techniques. For odd values of $n$, it is expressed in Bessel functions, while, for even values of $n$ it is expressed in terms of elementary functions. Thus, for $n=2$, one has $$ \int_{-1}^1 \cos(\alpha t)(1{-}t^2)^{(2-2)/2}dt = \frac{2\sin\alpha}{\alpha}, $$ while, for $n=3$, one has $$ \int_{-1}^1 \cos(\alpha t)(1{-}t^2)^{(3-2)/2}dt = \frac{\pi\ J_1(\alpha)}{\alpha}, $$ where $J_1$ denotes the Bessel function of the first kind, and so on.