# Integral convergence in scattering problem

Consider the scattering of a particle by a central time-independent potential $$V$$, that is limited to a finite region in the space. the hamiltonian is $$H = H_{0} + V$$. Using the Moller operators, we can express the state of the particle at time $$t = 0$$: $$|\psi \rangle = \Omega_{+} |\psi_{in} \rangle = \lim_{t \rightarrow -\infty} U^{\dagger}(t) U_{0}(t) |\psi_{in} \rangle,$$ where $$|\psi_{in} \rangle$$ is an asymptotic state and $$U(t) = e^{-iHt}$$, $$\ U_{0}(t) = e^{-iH_{0}t}$$ are time evolution operators. Deriving and integrating $$U^{\dagger}(t) U_{0}(t)$$, it can be shown that you can write $$|\psi \rangle = |\psi_{in} \rangle + i \int_{0}^{\infty} d\tau \ U^{\dagger}(\tau)V U_{0}(\tau) |\psi_{in} \rangle.$$ The last integral is convergent and absolutely convergent, so it can be repleced by $$\lim_{\epsilon \rightarrow 0^{+}}\int_{0}^{\infty} d\tau \ e^{-\epsilon \tau} U^{\dagger}(\tau)V U_{0}(\tau) |\psi_{in} \rangle.$$ My questions are:

1) How we know that this integral converge? it is not obvious to me.

2) What is the need to use this $$\epsilon$$? It always appear in scattering theory, why is it important?

Your potential function $$V$$ is time independent, and is limited to a finite region of space. That means that $$V\in[a,b]$$ where $$a\geq0$$, $$b\geq0$$. The result is that
$$|\psi \rangle = |\psi_{in} \rangle + i \int_{0}^{\infty} d\tau \ U^{\dagger}(\tau)V U_{0}(\tau) |\psi_{in} \rangle$$
$$|\psi \rangle = |\psi_{in} \rangle + i \int_{a}^{b} d\tau \ U^{\dagger}(\tau)V U_{0}(\tau) |\psi_{in} \rangle$$
because for all $$\tau\notin[a,b]$$ $$V=0$$. I make the assumption that none of the functions in your integrand are singular anywhere on the interval $$[a,b]$$. If this is all true, then the integrand is zero.