I am studying introductory quantum field theory and I am trying to reduce the integral over the relativistic phase space of three particles of arbitrary masses to an integral over two energies and three angles. From this source (page 3 below figure 49.2) I got the idea that perhaps the three angles could be the Euler angles. I am also looking at this source.
The integral I am talking about is the following:
$$ \int \frac{d^3\vec{p_1}}{(2\pi)^3 2 E_1}\frac{d^3\vec{p_2}}{(2\pi)^3 2 E_2}\frac{d^3\vec{p_3}}{(2\pi)^3 2 E_3} (2\pi)^4\delta^4(P - p_1 - p_2 - p_3) $$
My idea is the following: \begin{align*} &= \int \frac{d^3\vec{p_1}}{(2\pi)^3 2 E_1}\frac{d^3\vec{p_2}}{(2\pi)^3 2 E_2}\frac{d^3\vec{p_3}}{(2\pi)^3 2 E_3} (2\pi)^4\delta^4(P - p_1 - p_2 - p_3) \\ &= \int \frac{d^3\vec{p_1}}{(2\pi)^3 2 E_1}\frac{d^3\vec{p_2}}{(2\pi)^3 2 E_2}\int\frac{\vec{p_3} \cdot \vec{p_3} dp_3 d\Omega}{(2\pi)^3 2 \sqrt{\vec{p_3} \cdot \vec{p_3} + m^2}} (2\pi)^4\delta^4(P - p_1 - p_2 - p_3)\\ &= \int \frac{d^3\vec{p_1}}{(2\pi)^3 2 E_1}\frac{d^3\vec{p_2}}{(2\pi)^3 2 E_2}\int\frac{(P-p_1-p_2)^2 dp_3 d\Omega}{(2\pi)^3 2 \sqrt{(P-p_1-p_2)^2 + m^2}} (2\pi)^4\delta^4(P - p_1 - p_2 - p_3)\\ \end{align*}
I am not sure whether this is correct. How can I re-express the relativistic phase space integral over one involving two energies and three angles?
The usual trick is to using the formula $$\frac{1}{2E}=\int dp^0 \Theta(p^0)\delta(p^2-m^2), \quad E=\sqrt{\vec{p}^2+m^2}, \tag{1} \label{2}$$ where $\Theta(p^0)$ denotes the Heaviside step function. In this way, the three-particle phase space integral can be recast into the simpler form $$\begin{align}I_3 &=\int \frac{d^3p_1}{2E_1} \frac{d^3 p_2}{2E_2}\frac{d^3p_3}{2E_3} \,\delta^{(4)}(P-p_1-p_2-p_3) \ldots \\[5pt] &=\int \frac{d^3 p_1}{2E_1} \frac{d^3 p_2}{2 E_2} d^4 p_3 \, \Theta(p_3^0) \delta(p_3^2-m_3^2)\delta^{(4) }(P-p_1-p_2-p_3)\ldots \\[5pt] &= \int \frac{d^3 p_1}{2E_1} \frac{d^3p_2}{2E_2} \Theta(P^0-E_1-E_2) \delta\left[(P-p_1-p_2)^2-m_3^2 \right]\ldots,\tag{2} \end{align} $$ where the integration over the $4$-dimensional delta function was carried out in the last step.
As an example, let us consider the scalar integral $$I=\int \frac{d^3p_1}{2E_1} \frac{d^3p_2}{2E_2} \frac{d^3p_3}{2E_3} \delta^{(4)}(P-p_1-p_2-p_3) f(p_1\cdot p_2). \tag{3} \label{3}$$ As the integral $I$ is a Lorentz scalar, it can only depend on the Lorentz invariant $P^2$. As a consequence, the integration in \eqref{3} can be performed in any (inertial) reference frame. The most convenient choice is the center-of-mass frame, where $P^\mu =(P^0, \vec{P})=(\sqrt{P^2}, \vec{0})$. Working out the final result (with only a single integration left), $$\begin{align}I &= \frac{\pi^2}{4 P^2} \! \!\!\!\int\limits_{(m_1+m_2)^2}^{(\sqrt{P^2}-m_3)^2} \! \! \!\!\frac{ds}{s} w(s,m_1^2,m_2^2)w(P^2,m_3^2,s)f\left((s-m_1^2-m_2^2)/2 \right) \Theta(P^0), \tag{4} \label{4}\\[5pt] w(a,b,c)&=\sqrt{a^2+b^2+c^2-2ab-2ac-2bc}, \end{align}$$ is left as a homework exercise.
For further details see e.g. H. Pietschmann, Weak Interactions - Formulae, Results, and Derivations, Springer, 1983, pp. 19-23.