I have a question involving surface integral on a unit sphere. Suppose $s_1$ and $s_2$ are two points on a unit sphere with spherical coordinates $(\theta_1, \psi_1)$ and $(\theta_2, \psi_2)$, respectively. I want to compute
$$\int_{{\bf x}\in \mathbb{S}^2} \exp\{s^T_1 \cdot {\bf x}\} \exp\{s^T_2 \cdot {\bf x}\}d\mathbf{x}$$
where $\mathbb{S}^2$ stands for the unit sphere. $s_1^T \cdot \mathbf{x}$ means the dot product of the vectors $\vec{os_1}$ and $\vec{o\mathbf{x}}$ where $o$ is the center of the sphere.
Here is my idea. I can write $d\mathbf{x}$ as $\sin \theta d\theta d\psi$ where $(\theta, \psi)$ is the spherical coordinates of $\mathbf{x}$. Without loss of generality, I can assume $s_1$ is the top point of the sphere (on the $z$ axis) so that $s_1^T \cdot \mathbf{x}$ becomes $\cos \theta$. Now I think if I can write $s_2^T \cdot \mathbf{x}$ as a function of $\theta, \theta_2, \psi, \psi_2$, I can try to do this double integral. So my question boils down to "What is the form of $s_2^T \cdot \mathbf{x}$ in terms of their spherical coordinates $\theta, \theta_2, \psi, \psi_2$".
Maybe I can do this surface integral in an easier way. Any comments and suggestions are welcome.
Best Answer
Let \begin{equation} I=\int d\Omega\, e^{({\bf s}_1+{\bf s}_2)\cdot \hat{\bf r}}, \end{equation} where $d\Omega=\sin(\theta) d\phi d\theta$ and $\hat{\bf r}$ is a unit vector in spherical coordinates.
We are free to choose our coordinate system so, we choose the $\hat{\bf z}$ to lie along the ${\bf s}_1+{\bf s}_2$ direction. Then using $({\bf s}_1+{\bf s}_2)\cdot \hat{\bf r}=|{\bf s}_1+{\bf s}_2|\cos(\theta)$, where $|{\bf s}_1+{\bf s}_2|$ is the norm of the vector, \begin{equation} I=\int\limits_0^{2\pi}d\phi\int\limits^{\pi}_{0}d\theta\, \sin(\theta)e^{|{\bf s}_1+{\bf s}_2|\cos(\theta)}. \end{equation} The $\phi$ integral is trivial, while the $\theta$ integral can be easily done with the substitution $u=\cos(\theta)$ giving \begin{equation} I=4\pi\frac{\sinh(|{\bf s}_1+{\bf s}_2|)}{|{\bf s}_1+{\bf s}_2|}. \end{equation}