= Solution
Take an on-shell <W boson> of fixed <mass> $M_W$, zero transverse recoil, negligible <charged lepton>[charged-lepton] <mass>, and unrestricted angular acceptance. In its rest frame the charged <lepton> has <momentum> magnitude $p_0=M_W/2$. A longitudinal <Lorentz boost> to the laboratory preserves <transverse momentum>, so
$$
p_T=p_0\sin\theta^\star,\qquad
c=\pm\sqrt{1-(p_T/p_0)^2}.
$$
Both angular roots contribute to the same $p_T$. Set $x=p_T/p_0$ and $a=\sqrt{1-x^2}$. The change-of-variable <Jacobian determinant> has $|dc/dp_T|=x/(p_0a)$, and the two-root angular weight is $\tfrac38[(1-a)^2+(1+a)^2]=\tfrac34(2-x^2)$. Consequently
$$
\boxed{\frac1\Gamma\frac{d\Gamma}{dp_T}
=\frac{3}{4p_0}\frac{x(2-x^2)}{\sqrt{1-x^2}},\qquad
0<p_T<p_0,\qquad p_T^{\max}=\frac{M_W}{2}.}
$$
This is the ideal <W-decay Jacobian peak>. The divergence at the endpoint is integrable; it is a change-of-variable effect, not an infinite decay probability. Reversing the polar axis does not change this <transverse momentum> density, because the sum of the two roots removes the term odd in $c$.
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