# LSZ Reduction formula: Peskin and Schroeder

On page 224 of P&S, they have the following expression (7.36),

The integral over $$d^3q$$ gives us all the $$q \to p$$, then the integral over $$dx^0$$ is computed. The RHS given matches when only the $$x^0 = T_+$$ term of the integral is kept, what happens to the $$x^0=\infty$$ term?

$$\sum_{\lambda}\int_{T_+}^{\infty}dx^0\int\frac{d^3q}{(2\pi)^3} \frac{1}{2E_q(\lambda)} e^{i(p^0-q^0+i\epsilon)x^0}\langle\Omega|\phi(0)|\lambda_0\rangle (2\pi)^3\delta^3(\textbf{p}-\textbf{q}) \\\times\langle\lambda_{\textbf{q}}|T\{\phi(z_1)...\}|\Omega\rangle=\\ \sum_{\lambda}\int_{T_+}^{\infty}dx^0\int\frac{d^3q}{(2\pi)^3} \frac{1}{2E_q(\lambda)} \bigg[\frac{e^{i(p^0-q^0+i\epsilon)x^0}}{i(p^0-q^0+i\epsilon)}\bigg]_{T_+}^{\infty} \\\langle\Omega|\phi(0)|\lambda_0\rangle (2\pi)^3\delta^3(\textbf{p}-\textbf{q}) \times\langle\lambda_{\textbf{q}}|T\{\phi(z_1)...\}|\Omega\rangle$$ from which one only needs to consider the term in the brackets, namely $$\frac{e^{i(p^0-q^0+i\epsilon)x^0}}{i(p^0-q^0+i\epsilon)} \Bigg|_{T_+}^{\infty}= -i\lim_{x^0\to\infty} \frac{e^{i(p^0-q^0+i\epsilon)x^0}}{(p^0-q^0+i\epsilon)} +i\frac{e^{i(p^0-q^0+i\epsilon)T_+}}{(p^0-q^0+i\epsilon)}$$ The last term is the term you are looking for, see LHS of Eq. (7.36) in P&S. The first term vanishes, since $$-i\lim_{x^0\to\infty} \frac{e^{i(p^0-q^0+i\epsilon)x^0}}{(p^0-q^0+i\epsilon)}= -\frac{i}{(p^0-q^0+i\epsilon)} \lim_{x^0\to\infty}e^{i(p^0-q^0)x^0} \lim_{x^0\to\infty}e^{-\epsilon x^0}$$ This result is obviously zero, since the second limit vanishes, whereas the first one rapidly oscillates.