vanhees71 said:
I'd put ##u=4m^2-s-t##. Then the cross section depends only on ##s## and ##t##. As shown above through ##t## the expression depends on the scattering angle ##\vartheta##. Then you can check, where the expression at fixed ##s=E_{\text{cm}}^2## is maximal as a function of ##t## and thus which angles dominate.
Ah! I see!
So, you suggest the following:
##\displaystyle{\frac{d\sigma}{d\theta} = \frac{|\mathcal{M}|^{2}}{64\pi^{2}E_{\text{cm}}^{2}}}##
##\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left|\frac{i}{s-m^{2}} + \frac{i}{t-m^{2}} + \frac{i}{u-m^{2}}\right|^{2}}##
##\displaystyle{= \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{t-m^{2}} + \frac{1}{u-m^{2}}\right)^{2}}##
##\displaystyle{= \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{t-m^{2}} + \frac{1}{(4m^{2}-s-t)-m^{2}}\right)^{2}}##
##\displaystyle{= \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{t-m^{2}} + \frac{1}{3m^{2}-s-t}\right)^{2}}##
##\displaystyle{\frac{d\sigma}{d\theta} = \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{-\sqrt{s-4m^{2}}(1-\cos\theta)-m^{2}} + \frac{1}{3m^{2}-s+\sqrt{s-4m^{2}}(1-\cos\theta)}\right)^{2}}##
For ##s\gg \sqrt{m}##,
##\displaystyle{\frac{d\sigma}{d\theta} = \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s} + \frac{1}{-\sqrt{s}(1-\cos\theta)-m^{2}} + \frac{1}{-s+\sqrt{s}(1-\cos\theta)}\right)^{2}}##
So, the differential cross-section, for a fixed ##s##, is a maximum at angles ##\theta = 2\pi n##, where ##n=0, \pm 1, \pm 2, \dots##.
What do you think?