# Why do we use pole mass in LSZ formula?

I am reading "Quantum Field theory and the Standard Model" by Schwartz. It derives LSZ formula in chapter 6,

$$$$\langle f|S|i\rangle =\left[i\int d^4x_1e^{-ip_1x_1}(\Box + m^2)\right]...\left[i\int d^4x_ne^{-ip_nx_n}(\Box + m^2)\right]\times\langle\Omega|T\{\phi(x_1)\phi(x_2)\phi(x_3)...\phi(x_n)\}|\Omega\rangle$$$$

After that the book gives examples of tree-level calculations for scattering amplitude of some simple QED process. So far so good.

But then in later chapters the book talks about renormalization where I am told that the $$m$$ in Lagrangian isn't actually the pole mass, but is in fact infinite so it cancels some infinities from loop diagrams. I sort of understand how $$m$$ in Lagrangian isn't an observable so it doesn't have to be the pole mass, but then how do we know that the value of $$m$$ in LSZ formula must be pole mass? I re-read the derivation of LSZ formula in Schwartz and the only justification seems to be that at $$t=\pm\infty$$, the field $$\phi$$ satisfies the non-interacting Klein-Gordon equation, but doesn't the mass $$m$$ in Klein-Gordon equation comes from Lagrangian, which should be infinite?

Because in LSZ you have $$(\partial^2 + m^2)$$ and the propagator in momentum space is $$\frac{i}{-k^2 + m^2}$$ so you can see that when a plane wave acts on the differential operator you get something proportional to 1/propagator, so that will cancel out the propagators to get a finite result, which is only the case if the $$m$$ that shows in the differential operator is the pole $$m$$.