Show that
$$
2\left(k \cdot p^{\prime}\right)\left(k^{\prime} \cdot p\right)=\left(p-k^{\prime}\right)^2\left(k^{\prime} \cdot p\right)=\left(m^2-2 m \omega^{\prime}\right) m \omega^{\prime}
$$
in the muon rest frame, where $p=(m, 0,0,0)$.
Gathering these results together, the decay rate in the muon rest frame is
$$
\begin{aligned}
d \Gamma= & \frac{G^2}{2 m \pi^5} \frac{d^3 p^{\prime}}{2 E^{\prime}} \frac{d^3 k^{\prime}}{2 \omega^{\prime}} m \omega^{\prime}\left(m^2-2 m \omega^{\prime}\right) \\
& \times \delta\left(m^2-2 m E^{\prime}-2 m \omega^{\prime}+2 E^{\prime} \omega^{\prime}(1-\cos \theta)\right),
\end{aligned}
$$
and, as for $\beta$-decay, we can replace $d^3 p^{\prime} d^3 k^{\prime}$ by
$$
4 \pi E^{\prime 2} d E^{\prime} 2 \pi \omega^{\prime 2} d \omega^{\prime} d \cos \theta .
$$
We now use the fact that
$$
\delta\left(\cdots+2 E^{\prime} \omega^{\prime} \cos \theta\right)=\frac{1}{2 E^{\prime} \omega^{\prime}} \delta(\cdots-\cos \theta)
$$
to perform the integration over the opening angle $\theta$ between the emitted $\mathrm{e}^{-}$and $\bar{\nu}_e$ and obtain
$$
d \Gamma=\frac{G^2}{2 \pi^3} d E^{\prime} d \omega^{\prime} m \omega^{\prime}\left(m-2 \omega^{\prime}\right) .
$$
The $\delta$-function integration introduces the following restrictions on the energies $E^{\prime}, \omega^{\prime}$, stemming from the fact that $-1 \leq \cos \theta \leq 1$ :
$$
\begin{aligned}
\frac{1}{2} m-E^{\prime} & \leq \omega^{\prime} \\
0 & \leq \frac{1}{2} m, \\
0 E^{\prime} & \leq \frac{1}{2} m .
\end{aligned}
$$
These limits are easily understood in terms of the various limits in which the three-body decay $\mu \rightarrow \mathrm{e} \bar{\nu}_e \nu_\mu$ becomes effectively a two-body decay. For example, when the electron energy $E^{\prime}$ vanishes, $(12.39)$ yields $\omega^{\prime}=m / 2$, which is expected because then the two neutrinos share equally the muon's rest energy.
To obtain the energy spectrum of the emitted electron, we perform the $\omega^{\prime}$ integration of (12.38):
$$
\begin{aligned}
\frac{d \Gamma}{d E^{\prime}} & =\frac{m G^2}{2 \pi^3} \int_{\frac{1}{2} m-E^{\prime}}^{\frac{1}{2} m} d \omega^{\prime} \omega^{\prime}\left(m-2 \omega^{\prime}\right) \\
& =\frac{G^2}{12 \pi^3} m^2 E^{\prime 2}\left(3-\frac{4 E^{\prime}}{m}\right) .
\end{aligned}
$$
This prediction is in excellent agreement with the observed electron spectrum. Finally, we calculate the muon decay rate
$$
\Gamma \equiv \frac{1}{\tau}=\int_0^{m / 2} d E^{\prime} \frac{d \Gamma}{d E^{\prime}}=\frac{G^2 m^5}{192 \pi^3} .
$$
Inserting the measured muon lifetime $\tau=2.2 \times 10^{-6} \mathrm{sec}$, we can calculate the Fermi coupling $G$. We find
$$
G \sim 10^{-5} / m_N^2 .
$$
Comparison of the values of $G$ obtained in (12.24) and (12.43) supports the assertion that the weak coupling constant is the same for leptons and nucleons, and hence universal. It means that nuclear $\beta$-decay and the decay of the muon have the same physical origin. Indeed, when all corrections are taken into
account, $G_\beta$ and $G_\mu$ are found to be equal to within a few percent:
$$
\begin{aligned}
G_\mu & =(1.16632 \pm 0.00002) \times 10^{-5} \mathrm{GeV}^{-2}, \\
G_\beta & =(1.136 \pm 0.003) \times 10^{-5} \mathrm{GeV}^{-2}
\end{aligned}
$$