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In Schwartz' "Quantum Field Theory and the Standard Model", chapter 20, the author calculates the cross section for a process to produce a muon-antimuon pair as well as an extra photon emmitted from the final state. In order to regulate infared divergences, he introduces a ficticuous photon mass, and the variable $\beta = \frac{m_\gamma^2}{Q^2}$ where $Q^2 = (p_\mu + p_\bar{\mu} + p_\gamma)^2$.

He then defines new variables in terms of the relevant Mandelstam ones:

$$s = (p_{\mu} + p_{\bar{\mu}})^2 = Q^2(1-x_\gamma),\tag{20.37}$$

$$t = (p_{\mu} + p_{\gamma})^2 = Q^2(1-x_1),\tag{20.38}$$

$$u = (p_{\bar{\mu}} + p_{\gamma})^2 = Q^2(1-x_2),\tag{20.39}$$

These variables obey the constraints $$x_\mu + x_\bar{\mu} + x_\gamma = 2 - \beta,\tag{20.40}$$ $0 \le x_\gamma \le 1$, and $0 \le x_1, x_2 \le 1 - \beta$. All of this is given in the text directly.

He then chooses to integrate out $x_\gamma$ with the energy conserving delta function, setting it equal to $2 - \beta - x_1 - x_2$, leaving the phase space integral in the cross section as the integral over $x_1$ and $x_2$ subject to the contraints stated above.

The resulting integration bounds are quoted as

$0 \le x_1 \le 1-\beta$: This one is clear.

$1 - \beta - x_1 \le x_2 \le 1 - \frac{\beta}{1-x_1}$: The lower bound is clear to me, as it is ensuring that the sum of $x_1$ and $x_2$ is never less than $1-\beta$, as in this case $x_\gamma$ would have to exceed 1. However, I do not understand how to arrive at the upper bound here. For example, the point $x_1 = 1-\beta, x_2 = 1 -\beta, x_\gamma = \beta$ satisfies all the constraints, and yet it is not permitted by this upper bound. Moreover, I am confounded as to how a non-linear boundary has arised from the linear constraints. I can't understand why the upper bound for this integral is not also $1-\beta$: as this seems to satisfy the constraints to me.

In short, how did Schwartz arrive at these integration bounds?

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The integral over $z = \cos \theta$ is of the form $$ \int_{-1}^1 dz \delta \left( 2 - x_1 - x_2 - \sqrt{ x_1^2 + x_2^2 + 2 x_1 x_2 z + 4 \beta } \right) $$ The argument of the delta function vanishes if $x_1 + x_2 \leq 2$ and $$ z = 1 + 2 \frac{1-\beta-x_1-x_2}{x_1x_2} $$ However, we get a non-zero contribution only if this value lies in $[-1,1]$. This implies $$ - 1 \leq 1 + 2 \frac{1-\beta-x_1-x_2}{x_1x_2} \leq 1 $$ Since $x_1,x_2>0$, the upper bound gives $1-\beta \leq x_1+x_2 \leq 2 $. The lower bound gives $$ x_2 \leq 1-\frac{\beta}{1 - x_1} $$ This is where your upper bound comes from. This bound is violated if you choose $x_1=x_2=1-\beta$.

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