I've worked something out, but I'm not sure if it's right. Here it is, just in case
Electrons can escape only if their velocity in the $ x $-direction is large enough to overcome the work function
$$ v_{x,\text{min}} = \sqrt{\frac{2 W}{m_e}} \tag{1} $$
and after crossing it, their kinetic energy in the $ x $-direction decreases by $ W $
$$ \frac{1}{2} m_e v_x'^2 = \frac{1}{2} m_e v_x^2 - W \tag{2} $$
Now to compute the Richardson-Dushman current, we integrate the velocity distribution inside over all electrons with
$ v_x \ge v_{x,\text{min}} $ :
$$ J = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{ v_{x,\text{min}} }^{\infty} e v_x n(\vec{v}) dv_x dv_y dv_z \tag{3} $$
and by changing variables using $ v_x' dv_x' = v_x dv_x $ the integral becomes
$$ J = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{ 0 }^{\infty} e v_x' n(\vec{v}) dv_x' dv_y dv_z \tag{4} $$
Since current is the same after crossing, we can also write it using the velocity distribution $ n' $ outside
$$ J = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{ 0 }^{\infty} e v_x' n'(\vec{v}') dv_x' dv_y dv_z \tag{5} $$
Implying:
$$ n'(\vec{v}') = n(\vec{v}) \tag{6} $$
For high energies, the distribution inside approximates a Maxwell-Boltzmann distribution :
$$ n'(\vec{v}') = \frac{2 m_e^3 }{h^{3}} \exp\left( - \frac{m_e(v_x^2 + v_y^2 + v_z^2)}{2kT} + \frac{\mu}{kT} \right) \tag{7} $$
where we can substitute in (2) to get
$$ n'(\vec{v}') = \frac{2 m_e^3 }{h^{3}} \exp \left( \frac{\mu - W}{k T} \right) \exp\left( - \frac{m_e(v_x'^2 + v_y^2 + v_z^2)}{2kT} \right) \tag{7} $$
