The question isn't about any actual homework, it's rather a (probably simple) intermediate step I've encountered on Rajaraman's Solitons and instantons : an introduction to solitons and instantons in quantum field theory, in the context of the Sine-Gordon equation. The full solution of the equation is a rather complicated problem, so we limit ourselves to particular solutions, one of which is the soliton-antisoliton scattering solution, which has the form: \begin{equation} \phi(x,t)=4arctan\left( \frac{sinh(ut/\sqrt{1-v^2})}{u \ cosh(x/\sqrt{1-v^2})}\right) \end{equation} He argues that in the limit that t goes to minus infinity, for example, this becomes \begin{equation} \phi\rightarrow 4arctan\left[exp\left(\frac{x+v(t+\Delta/2)}{\sqrt{1-v^2}}\right) \right] - 4arctan\left[exp\left(\frac{x-v(t+\Delta/2)}{\sqrt{1-v^2}}\right) \right] \end{equation} where \begin{equation} \Delta\equiv \frac{1-v^2}{v}lnv \end{equation} and a similar solution for the positive infinity case(Page 38 of Rajaraman's forementioned book). I tried to put the solution in a form in which I can use the arctangent addiction formula, but thus far no success. Closer I got was \begin{equation} 4arctan\left \{\left[\frac{exp\left(x + \gamma v(t+(lnv)/(v\gamma))\right)}{1+e^{2\gamma x}} \right] - \left[\frac{exp\left(x - \gamma v(t-(lnv)/(v\gamma))\right)}{1+e^{2\gamma x}} \right] \right \} \end{equation} Since we're studying an asymptotic case, I don't really understand why the first term on this last equation should even be there instead of going to zero.
I'm probably missing something pretty obvious, but whatever the reason is, I just don't get it.Thanks in advance for any help.