I am reading a paper and trying to solve collision integral. In Appendix A, there is an integral, $$ I^{(n)} = \int \frac{p^2_1\,dp_1 \,p^2_3\,dp_3\,p^2_4\,dp_4} {2E_12E_32E_4} 2\pi \delta (p_1 − p_3 − p_4)e^{−E_1/T} I_\Omega,$$ where \begin{align} I_\Omega &=\left( 4\pi \right)^3 \int d^3\lambda \, \frac{\sin p_1\lambda }{p_1\lambda}\frac{\sin p_3\lambda }{p_3\lambda}\frac{\sin p_4\lambda }{p_4\lambda} \\ &=\dfrac{32\pi^5}{p_1p_3p_4}\Bigg[\dfrac{\left( p_1-p_3+p_4\right) }{\left| p_1-p_3+p_4\right|}+\frac{(p_1+p_3-p_4) }{| p_1+p_3-p_4|}-\frac{(p_1-p_3-p_4) }{|p_1-p_3-p_4|}-\frac{(p_1+p_3+p_4) }{|p_1+p_3+p_4|}\Bigg]. \end{align}
They get a result $$I^{(n)}=\int \frac{16\pi^6 p_1^2}{E_1}e^{-E_1/T} \, dp_1$$ after integrating out dirac delta. I can't seem to understand how do they remove both $p_3$ and $p_4$ using one delta integral. Also, since all the terms in brackets of $I_\Omega$ are positive, it seems that term goes to zero. Can anyone help?
Note: $E_i$ and $p_i$ are energy and momentum of $i^{th}$ particle and here it is assumed that $m_3 =0$ and $m_4=0$ hence $E_3 = \sqrt{p_{3}^{2}+m_{3}^{2}}=p_{3} $ & similarly $E_4 = p_4$.
First, it should be noted that $p_1,p_3,p_4$ are limited to positive real values (they represent magnitudes of momenta). Since the delta function restricts to $p_1=p_3+p_4$, it follows that
$$p_1-p_3+p_4=2p_4\geq 0,\quad p_1+p_3-p_4=2p_3\geq 0,\quad p_1+p_3+p_4=2p_1\geq 0.$$
Therefore for three of the four terms in $I_\Omega$ we have $$\dfrac{p_1-p_3+p_4}{| p_1-p_3+p_4|}=\dfrac{p_1+p_3-p_4}{| p_1+p_3-p_4|}=\dfrac{p_1+p_3+p_4}{| p_1+p_3+p_4|}=1.$$
This leaves the third term in $I_\Omega$, which at first glance seems ill-defined since $p_1-p_3-p_4=0$. But this absolute value arises from the representation $\int\frac{d\lambda}{\lambda p}\sin(\lambda p)=|p|^{-1},$ and thus $$\frac{p}{|p|}=\int_0^\infty \frac{d\lambda}{\lambda}\sin(\lambda p)\to \int_0^\infty \frac{d\lambda}{\lambda}0=0$$ in the limit $p\to 0^+$.
Combining the above, we see that $I_\Omega=\dfrac{32\pi^5}{p_1p_3p_4}$ on the restriction $p_1=p_3+p_4$. Hence
\begin{align} I^{(n)} &= \int \frac{p^2_1\,dp_1}{2E_1} \frac{p^2_3\,dp_3}{2E_3} \frac{p^2_4\,dp_4}{2E_4} 2\pi \delta (p_1 − p_3 − p_4)e^{−E_1/T} I_\Omega\\ &= \int \frac{p^2_1\,dp_1}{2E_1} \frac{p^2_3\,dp_3}{2E_3} \frac{p^2_4\,dp_4}{2E_4} 2\pi \delta (p_1 − p_3 − p_4) \dfrac{32\pi^5}{p_1p_3p_4}e^{−E_1/T}\\ &= \int \frac{8\pi^6 p_1}{E_1}e^{−E_1/T}\,dp_1 \,dp_3\,dp_4 \delta (p_1 − p_3 − p_4) \end{align} where I've used $E_3=p_3$, $E_4=p_4$. Eliminating the $p_4$-integral using the delta function yields
$$I^{(n)}=\int_0^\infty \frac{8\pi^6 p_1}{E_1}e^{−E_1/T}\,dp_1\int_0^{p_1}\,dp_3=\int_0^\infty \frac{8\pi^6 p_1^2}{E_1}e^{−E_1/T}\,dp_1.$$ This is almost in agreement with the stated result: there's a missing factor of two which I haven't tracked down yet. I'll see what I find.
I haven't succeeded in finding the error, and the more checks I do make me suspicious about the original source. In particular, the factor of two would be accounted for if one assumes that $$\frac{p_1-p_3-p_4}{|p_1-p_3-p_4|}\to -1$$ when $p_1=p_3+p_4$ rather than zero (as I have above). But as far as I can tell, this term should genuinely vanish given its integral representation in the Appendix. Beyond that I don't see an obvious explanation, and so I would tentatively propose a mathematical error in the source.
After having examined the source material (along with some awareness of the underlying physics) the current conjecture of the OP and myself is that the delta function should actually be $$\delta(E_1-E_3-E_4)=0.$$ This reflects the fact that the constraint should arise from conservation of 4-momentum, which includes energy conservation. With this reading in mind, the behavior of $I_\Omega(p_1,p_3,p_4)$ when subject to this constraint is a bit different and I'll address the terms in pairs. First, note that energy conservation implies $$p_3+p_4=E_3+E_4=E_1=\sqrt{p_1^2+m_1^2}\geq p_1 >0.$$ Since $p_1\geq 0 \geq -p_1$, we conclude that $p_1-p_3-p_4$ and $p_1+p_3+p_4$ have opposite signs. As such, the last two terms in $I_\Omega$ will vanish identically. The analogous question for the remaining two terms are the values of $p_1-p_3+p_4$ and $p_1+p_3-p_4$, with $I_\Omega$ vanishing unless the signs match. But $$(p_1-p_3+p_4)+(p_1+p_3-p_4)=2p_1\geq 0$$ and so cannot both be negative. For both to be positive, we require \begin{align} 0 &<(p_1-p_3+p_4)(p_1+p_3-p_4)\\ &=p_1^2-(p_3-p_4)^2\\ &=(p_3+p_4)^2-m_1^2-(p_3-p_4)^2\\ &=4p_3 p_4-m_1^2 \end{align} Thus the terms are both positive if $4p_3p_4>m_1^2$ and we conclude that energy conservation dictates $$I_\Omega(p_1,p_3,p_4)=\dfrac{64\pi^5}{p_1p_3p_4}\theta(4p_3p_4-m_1^2)$$ whenever energy conservation is fulfilled. As such, the complete integral is
$$I^{(n)} = \int \frac{p^2_1\,dp_1}{2E_1} \frac{p^2_3\,dp_3}{2E_3} \frac{p^2_4\,dp_4}{2E_4} 2\pi \delta (E_1 − E_3 − E_4)e^{−E_1/T}\dfrac{64\pi^5}{p_1p_3p_4}\theta(4p_3p_4-m_1^2)$$
Since energy conservation dictates $$p_4=E_4=E_1-E_3=\sqrt{p_1^2+m_1^2}-p_3,$$ integrating out $p_4$ first yields
$$I^{(n)} = \int \frac{16\pi^6 p_1\,dp_1}{E_1} dp_3 e^{−E_1/T}\theta\Bigg(4p_3\Big[\sqrt{p_1^2+m_1^2}-p_3\Big]-m_1^2\Bigg)$$
The limits for the $p_3$ integral are dictated by the condition
$$4p_3\Big[E_1-p_3\Big]-m_1^2\geq 0$$
which for $p_1,p_3\geq 0$ is equivalent to $$\frac12\left(E_1-p_1\right)\leq p_3\leq \frac12\left(E_1+p_1\right)$$ As such the $p_3$-integral is simply
$$\int_{(E_1-p_1)/2}^{(E_1+p_1)/2} dp_3 = \frac{E_1+p_1}{2}-\frac{E_1-p_1}{2}=p_1$$ and therefore $$I^{(n)} = \int \frac{16\pi^6 p_1^2\,dp_1}{E_1} e^{−E_1/T}$$ as claimed. As a check we consider the integral denoted $I^{(p)}$ in the source material. Then the presence of the factor $U=E_3=p_3$ renders the $p_3$ integral as
$$\int_{(E_1-p_1)/2}^{(E_1+p_1)/2} p_3\,dp_3 = \frac12\left(\frac{E_1+p_1}{2}\right)^2-\frac12\left(\frac{E_1-p_1}{2}\right)^2=\frac12 E_1 p_1$$ and therefore
$$I^{(p)} = \int 8\pi^6 p_1^2\,dp_1 e^{−E_1/T}$$ which again matches the source material.