= Solution
Substitute the preceding <spin average> into the supplied <differential scattering cross-section>. Since $e^4=16\pi^2\alpha^2$,
$$
\frac{d\sigma}{d\Omega}=\frac{\alpha^2}{4s}(1+\cos^2\theta).
$$
The angular integral is $2\pi\int_{-1}^1(1+x^2)\,dx=16\pi/3$. Consequently the <relativistic scattering cross-section> for the massless <electron-positron annihilation into a muon pair> is
$$
\boxed{\sigma=\frac{4\pi\alpha^2}{3s}.}
$$
The outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form gives
$$
\boxed{B=\frac{4\pi}{3},\qquad j=2,\quad k=-1,\quad l=0,\quad m=0.}
$$
The integrated answer depends only on the incoming invariant $s$; the angular variables $t,u$ have been integrated out.
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