# Analytic continuation of Feynman amplitudes seems ill-defined

I was reading Peskin & Schroeder's book on Quantum Field Theory and on chapter 7, "The optical theorem for Feynman diagrams" (page 232) they extend analytically the Feynman amplitude $$i \mathcal{M}(s)$$ to include complex values of $$s$$. He then says that physical amplitudes should be those who have a small positive complex part, meaning $$i \mathcal{M}(s + i\eta)$$. Also, now that we have extended the function we are interested in the discontinuity

$$i \mathcal{M}(s + i\eta) - i \mathcal{M}(s - i\eta).$$

When he calculates the $$s$$-channel one-loop diagram in $$\frac{\lambda}{4!} \phi^4$$ theory he calculates the diagram as the integral:

$$i \delta\mathcal{M}(s) = \frac{\lambda^2}{2} \int \frac{d^4 q}{(2\pi)^4} \frac{1}{(k/2-q)^2-m^2+i\epsilon}\frac{1}{(k/2+q)^2-m^2+i\epsilon} . \tag{1}$$

The poles of the function inside the integral are:

$$q_0 = \frac{1}{2}k_0 \pm (E_q - i\epsilon) \quad\text{and}\quad q_0 = - \frac{1}{2}k_0 \pm (E_q - i\epsilon) .$$

Doing all the math he gets to the result:

$$i \delta\mathcal{M}(s) = - 2 \pi i \frac{\lambda^2}{2} \frac{4 \pi}{(2\pi)^4} \int_{m}^{+\infty} dE_q E_q \lvert \mathbf{q} \rvert \frac{1}{2E_q} \frac{1}{k_0 (k_0 - 2E_q +i \epsilon)} . \tag{2}$$

He doesn't write the $$+i\epsilon$$ but I figured it must be there.

Only now does he make the substitution $$k_0 \to k_0 + i \eta$$, but it seems to me that this is completely arbitrary and the step of the calculation where you decide to make this substitution changes the result of the integral. For example, if we had done this before calculating the very first integral (1), the poles would have shifted of a quantity $$\pm i \eta$$, which would have to be compared to $$\pm i\epsilon$$ (and how do you even compare these two infinitesimals?) . This could shift the poles below or above the real line, essentially changing the result of the first integral. Also, Peskin & Schroeder ignoring the $$+i\epsilon$$ in the final result (2), and calling both $$\eta$$ and $$\epsilon$$, the same way (just $$\epsilon$$) seems to me like a way of not dealing with having to compare the two infinitesimals $$\eta$$ and $$\epsilon$$. I don't understand how to fix this.