Find average over surface of unit sphere $$< \exp(\vec b \cdot \vec n) (\vec a \cdot \vec n) > = \int_{S^2}  \exp(\vec b \cdot \vec n) (\vec a \cdot \vec n) dS$$
I have tried to add parmetr under the exponet:
$$I(S, t) = \int_{S^2}  \exp(t \vec b \cdot \vec n) (\vec a \cdot \vec n) dS$$
$$\cfrac{dI(S, t)}{dt} = \int_{S^2}  \exp(t \vec b \cdot \vec n) (\vec b \cdot \vec n) (\vec a \cdot \vec n) dS$$
But it did not simplify calculation. Anoter way is to orient z axis orthogonally to $\vec b$ and $\vec a$, and orient x axis along the $\vec b$, then ($\alpha$ is angle between $\vec b$ and $\vec a$):
$$< \exp(\vec b \cdot \vec n) (\vec a \cdot \vec n) > = \int_0^{2\pi} \int_0^{\pi} \exp(b \sin(\theta) \cos(\phi)) a \sin(\theta) \sin(\phi - \alpha) \sin(\theta) d\theta d\phi$$
But i can't to solve it.
 A: We can choose the polar system of coordinates and direct the polar axis $Z$ along the vector $\vec b$. If we denote $(\phi, \theta)$ the angles of the vector $\vec n$ in this system, and $(\phi_1, \theta_1)$ the angles of the vector $\vec a$, we get:

*

*$(\vec b, \vec n)=b\cos\theta$

*$(\vec a, \vec n)=a\big(\sin\theta_1\sin\theta\cos\phi_1\cos\phi+\sin\theta_1\sin\theta\sin\phi_1\sin\phi+\cos\theta_1\cos\theta\big)$
Integrating in the polar system with respect to $\phi$ (from $0$ to $2\pi$) we notice that two first terms in 2. are dying out. Therefore, we are left with
$$I=\int_{S^2}  \exp(\vec b \cdot \vec n) (\vec a \cdot \vec n) dS=2\pi a\cos\theta_1\int_0^\pi d\theta\,\sin\theta \,e^{b\cos\theta}\cos\theta$$
$$=2\pi a\cos\theta_1\int_{-1}^1 e^{bx}xdx=2\pi a\cos\theta_1\frac{\partial}{\partial b}\int_{-1}^1 e^{bx}dx=4\pi a\cos\theta_1\frac{\partial}{\partial b}\frac{\sinh b}{b}$$
$$I=4\pi a\cos\theta_1\frac{b\cosh b-\sinh b}{b^2}=4\pi (\vec a,\vec b)\frac{b\cosh b-\sinh b}{b^3}$$
