Spherical integral

Question

Let $y \in \mathbb{R}^n$ be fixed. Is there a nice expression for the following integral taken over the unit sphere in $\mathbb{R}^n$? $$ \int_{\|x\|=1} e^{2\pi i (x \cdot y)}~dx $$

Answer 1

First, consider the $n=3$ case to have a point of reference. Since we integrate over all directions on the sphere, we may take $\vec{y}$ to define the vertical axis i.e. $\vec{x}\cdot\vec{y}=y \cos\theta$ where $\theta$ is the azimuthal angle. Then the integral in spherical coordinates (normalized by the surface area) is

\begin{align}\frac{1}{4\pi}\int_{\Vert x\Vert=1}\exp({2\pi i\, \vec{x}\cdot \vec{y}})\,d^2x &=\frac{1}{4\pi}\int_0^{\pi}\!\!\int_0^{2\pi}\exp({2\pi i\, y \, \cos\theta})\sin\theta\,d\phi\,d\theta\\ &=\frac{1}{2}\left[-\frac{\exp({2\pi i y \cos\theta})}{2\pi i y}\right]_0^\pi\\ &=\frac{1}{4\pi iy}\left(e^{2\pi i y}-e^{-2\pi i y}\right)=\frac{\sin(2\pi y)}{2\pi y} \end{align}

Note that this goes to $1$ as $y\to 0$ as it should. (A reader versed in special functions will recognize this as the zeroth spherical Bessel function $j_0(2\pi y)$.)

We now want to generalize to arbitrary $n\geq 2$ i.e. the integral $$I_n(y):=\frac{1}{S_{n-1}}\int_{\Vert x\Vert=1}\exp{(2\pi i\,\vec{x}\cdot\vec{y})}\,d^{n-1}x.$$ Note that $I_n$ can only be a function of $y$ alone; also, this integral is normalized as $I_n(y=0)=1$. We can show that $I_n(y)$ satisfies a differential equation by taking the Laplacian of both sides:

\begin{align} \nabla^2 I_n(y) &=\frac{1}{y^{n-1}}\frac{d}{dy}\left( y^{n-1}\frac{d I_n}{dy}\right)\\ &=\sum_{k=1}^{n-1}\frac{\partial^2}{\partial x_k^2}\left[\frac{1}{S_{n-1}}\int_{\Vert x\Vert=1}\exp{(2\pi i\,\vec{x}\cdot\vec{y})}\,d^{n-1}x\right]\\ &=\frac{(2\pi i)^2}{S_{n-1}}\int_{\Vert x\Vert=1}\left(\sum_{k=1}^{n-1}x_k^2\right)\exp{(2\pi i\,\vec{x}\cdot\vec{y})}\,d^{n-1}x\\ &=\frac{(2\pi i)^2}{S_{n-1}}\int_{\Vert x\Vert=1}\exp{(2\pi i\,\vec{x}\cdot\vec{y})}\,d^{n-1}x &\left(\left(\sum_{k=1}^{n-1}x_k^2\right)=\Vert x \Vert^2=1\right)\\ &=-(2\pi)^2 I_n(y) \end{align}

To simplify the LHS we have recalled the form of the Laplacian in hyperspherical coordinates for the case of spherically symmetric potentials. So all that remains is to solve the linear second-order ODE above, which can be simplified to $$I_n''(y)+\left(\frac{n-1}{y}\right)I_n'(y)+(2\pi)^2 I_n(y)=0$$ This can be placed into the standard form of Bessel's equation by a suitable transformation. Rather than carry this out by hand, I'll just cite WolframAlpha:

$$I_n(y)=c_1 y^{1-n/2} J_{\frac{n}{2}-1}(2\pi y)+c_2 y^{1-n/2} Y_{\frac{n}{2}-1}(2\pi y)$$ where $J_n$ and $Y_n$ are $n$-th Bessel functions of the first and second kind respectively. We can ignore the second solution since it diverges at $y=0$; by comparison, the first solution may be checked to behave as $\pi^{n/2-1}/\Gamma(n/2) $ as $y\to 0$. Since $I_n(0)=1$ we conclude that $$I_n(y)=\frac{\Gamma(n/2)}{(\pi y)^{n/2-1}}J_{\frac{n}{2}-1}(2\pi y).$$ As a check, note that for $n=3$ we recover the same answer of $\dfrac{\sin 2\pi y}{2\pi y}$ as obtained earlier. (The case of odd $n$ gives Bessel functions of half-integer order i.e. spherical Bessel functions which can always be expressed through trig. functions.)

Since the $(n-1)$-dimensional surface area is $S_{n-1}=\pi^{n/2}/\Gamma\left(\frac{n}{2}+1\right)$, we conclude that

$$\int_{\Vert x\Vert=1}\exp{(2\pi i\,\vec{x}\cdot\vec{y})}\,d^{n-1}x = \frac{n \pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)} \frac{\Gamma(\frac n2)}{(\pi y)^{n/2-1}}J_{\frac{n}{2}-1}(2\pi y)=2\pi\,y^{1-n/2} J_{\frac{n}{2}-1}(2\pi y).$$