In Lancaster & Blundell's QFT book they show that
\begin{equation}A:= \int_{-\infty}^{\infty}dp \ p e^{ipx}e^{-it\sqrt{p^2+m^2}}\end{equation}
returns a nonzero value for $x$, $t$ and $m$ beeing positive real numbers.
First the branch points at $p=\pm i m$ are detected. Then the branch cuts and the contour are made as shown in the picture.
There is no pole inside of our integration courve, so the complete integral will return zero and therefore $-A$ must be equal to the rest of the contour integral.
Then they state, that with joran's lemma we see that the half-circle does not contribute to the integral. But i dont see it. For jordans lemma to work we first have to show that $p e^{-it \sqrt{p^2+m^2}}$ goes to zero as the radius of $p$ diverges on the upper half of the complex plane. But how do we show this?
Then for the remaining integral by substituting $p=iz$ they find \begin{equation} A=\int_m ^\infty d(iz) \ ize^{-zx}\left( e^{t\sqrt{z^2-m^2}}-e^{-t\sqrt{z^2-m^2}}\right). \end{equation}
My question is: Why?! Can somebody help me reproduce this in detail?
To prevent a wrong image: There are some arguments in the textbook, but i can't reproduce them, and i guess that simply copy & pasting them would hurt copyright.