Weinberg soft photon integral

Question

In deriving the rate of emission of arbitrary numbers of soft photons in a general QED process, Weinberg performs the following integral (equations 13.2.8-9):

$$-\pi(\vec{p}_m\cdot \vec{p}_n)\int_{\lambda\leq|\vec{q}|\leq\Lambda}\frac{d^3\vec{q}}{|\vec{q}|^3(E_n-\hat{q}\cdot\vec{p}_n)(E_m-\hat{q}\cdot\vec{p}_m)}=\frac{2\pi^2}{\beta_{nm}}\ln\left(\frac{1+\beta_{nm}}{1-\beta_{nm}}\right)\ln\left(\frac{\Lambda}{\lambda}\right)$$

where $$\beta=\sqrt{1-\frac{m_n^2m_m^2}{(\vec{p}_n\cdot\vec{p}_m)^2}}.$$

I am trying to compute this integral for myself but am having some trouble with the angular integral. Would someone mind giving me some assistance?

Answer 1

I think I solved it; unfortunately I don't have time to do all the manipulations after the integral is computed.

Let's change to spherical coordinates $$I=-\pi(\vec{p}_m\cdot\vec{p}_n)\int \text{d}q\text{d}\phi \sin(\phi) \text{d}\theta\frac{1}{q}\frac{1}{(E_n-p_n \cos(\phi))(E_m-p_m \cos(\phi))}$$

It follows that: $$I=-\pi(\vec{p}_m\cdot\vec{p}_n)\ln\left(\frac{\Lambda}{\lambda}\right)2\pi \int_0^{\pi}\text{d}\phi \frac{\sin(\phi)}{(E_n-p_n \cos(\phi))(E_m-p_m \cos(\phi))}$$

Now, let's change variables $$x=\cos(\phi) \\ \text{d}x=-\sin(\phi)\text{d}\phi$$ so we have $$I=-2\pi^2(\vec{p}_m\cdot\vec{p}_n)\ln\left(\frac{\Lambda}{\lambda}\right) \int_{-1}^{1}\text{d}x\frac{1}{(E_n-p_n x)(E_m-p_m x)}$$

Now, we have to make the following step: $$\frac{A}{(E_n-p_n x)}+\frac{B}{(E_m-p_m x)}$$ you find that $$A=\frac{-p_n}{E_n p_m-E_m p_n} \\ B=\frac{p_m}{E_np_m-E_mp_n}$$

Then, it's easy to see that $$I=2\pi^2(\vec{p}_m\cdot\vec{p}_n)\ln\left(\frac{\Lambda}{\lambda}\right)\left[\frac{A}{p_n}\ln\left(E_n-p_nx\right)+\frac{B}{p_m}\ln\left(E_m-p_mx\right)\right]^1_{-1}$$ by making $A$ and $B$ explicit we get: $$ I=2\pi^2\frac{(\vec{p}_m\cdot\vec{p}_n)}{E_np_m-E_np_n}\ln\left(\frac{\Lambda}{\lambda}\right) \left[-\ln\left(E_n-p_nx\right)+\ln\left(E_m-p_mx\right)\right]^1_{-1}$$ therefore $$I=2\pi^2\frac{(\vec{p}_m\cdot\vec{p}_n)}{E_np_m-E_np_n}\ln\left(\frac{\Lambda}{\lambda}\right)\left[\ln\left(\frac{E_m-p_m}{E_n-p_n}\right)+\ln\left(\frac{E_n+p_n}{E_m+p_m}\right)\right]$$ and then it's should just be rearranging the factors in a convenient way. I hope that helped!