# Expression in John R. Taylor Scattering theory book

In the book Scattering theory by John R. Taylor He has the following expression at page 138

$$i \lim _{\epsilon \downarrow 0} \int_{0}^{\infty} d t\left\langle\mathbf{p}^{\prime}\right|V e^{i\left(E_{p}^{\prime}+E_{p}+i \epsilon-2 H\right) t}|\mathbf{p}\rangle=1 / 2 \lim _{\epsilon \downarrow 0}\left\langle\mathbf{p}^{\prime}\right|V G\left(\frac{E_{p^{\prime}}+E_{p}}{2}+i \epsilon\right)|\mathbf{p}\rangle \tag 1$$

where $$G$$ is the green operator defined by $$G(z)=(z-H)^{-1}$$

But for me we should have $$i \lim _{\epsilon \downarrow 0} \int_{0}^{\infty} d t\left\langle\mathbf{p}^{\prime}\right|V e^{i\left(E_{p}^{\prime}+E_{p}+i \epsilon-2 H\right) t}|\mathbf{p}\rangle=1 / 2 \lim _{\epsilon \downarrow 0} \left\langle\mathbf{p}^{\prime}\right|V G\left(\frac{E_{p^{\prime}}+E_{p}+i \epsilon}{2}\right)|\mathbf{p}\rangle \tag 2$$ Since we have that

$$\int dt e^{i\left(E_{p}^{\prime}+E_{p}+i \epsilon-2 H\right) t} =\int dt e^{2i\left(\frac{E_{p^{\prime}}+E_{p}+i \epsilon}{2}+ H\right) t} =\frac{1}{2i\left(\frac{E_{p^{\prime}}+E_{p}+i \epsilon}{2}+ H\right)}e^{2i\left(\frac{E_{p^{\prime}}+E_{p}+i \epsilon}{2}+ H\right) t} \tag 3$$ And from $$(3)$$ we have expression $$(2)$$. I am not seeing how did he obtain expression $$1$$

Since $$\epsilon$$ is a small parameter that is taken to zero at the end of the calculation, and no physical quantities depend on $$\epsilon$$, it is conventional to "absorb" constant multiplicative factors into $$\epsilon$$. In your example, you could define $$\epsilon'\equiv \epsilon/2$$ and carry on with the calculation in terms of $$\epsilon'$$ instead of $$\epsilon$$. To save spending mental energy on a detail that is really irrelevant to a quite complex calculation, the convention is not to bother distinguishing $$\epsilon'$$ from $$\epsilon$$, and to refer to any $$\alpha \epsilon$$ (where $$\alpha$$ is a constant, positive factor) as $$\epsilon$$.