I saw a formula in this paper A. D. Becke (1983). Hartree–Fock exchange energy of an inhomogeneous electron gas. which is an integral about the spherical means: $$ \frac{1}{4\pi} \int e^{\vec{s}\cdot\nabla_1} \,d\Omega = \frac{\sinh (s\nabla_1)}{(s\nabla_1)} \tag1 $$ My equation is how to derive the result of this integral, shown on the right-handed sided in (1)?
Explanation:
$d\Omega$ is the differential of solid angle, $d\Omega = \sin \theta d \theta d \varphi$;
$e^{\vec{s} \cdot \nabla_1}$ should be understood as an abbreviation from the Taylor expansion, $$ e^{\vec{s} \cdot \nabla_1} = 1 + (\vec{s} \cdot \nabla_1) + \frac{1}{2!}(\vec{s} \cdot \nabla_1)^2 + \frac{1}{3!}(\vec{s} \cdot \nabla_1)^3 + \cdots $$ since the formula (11) in that paper, $\rho_{x\uparrow}(\vec{r}, \vec{r}+\vec{s}) = e^{\vec{s} \cdot \nabla_1}\, \rho_{x\uparrow}(\vec{r},\vec{r}_1)\big|_{\vec{r}_1 = \vec{r}}$, is apparently the Taylor Expansion in 3D, noting $\vec{s}$ is the small quantity for this expansion;
The notation $\nabla_1$ refers to act on the coordinates $r_1$.
The part about the integral in that paper is put as following:
In your equation $$ \frac{1}{4\pi} \int e^{\vec{s}\cdot\nabla_1} \,d\Omega = \frac{\sinh (s\nabla_1)}{(s\nabla_1)} \tag1 $$ just use the coordinate system where $\theta$ is the angle between $\vec s$ and $\nabla_1$. Then $$\vec{s}\cdot\nabla_1=s\nabla_1\cos\theta$$ This does not depend on $\varphi$, so $$\int_0^{2\pi}d\varphi=2\pi$$ Then using the substitution $x=\cos\theta$, $dx=-\sin\theta d\theta$, with the limits from $1$ to $-1$, you get $$\int_{-1}^1e^{\alpha x}dx=\frac1\alpha e^{\alpha x}\bigg|_{-1}^1=\frac2\alpha\sinh(\alpha)$$ I've used $\alpha=s\nabla_1$.