# Moving the integral $Q(x) = -\frac{e^{-1/2x}}{4i}\int_{1/2-i\infty}^{1/2+i\infty} \zeta(s)\Gamma(\frac{s}{2})\pi^{-s/2}e^{xs} ds$ past Re(s) = 1.

Given the integral $$Q(x) = -\frac{e^{-1/2x}}{4i}\int_{1/2-i\infty}^{1/2+i\infty} \zeta(s)\Gamma(\frac{s}{2})\pi^{-s/2}e^{xs} ds,$$

I know that the integrand is holomorphic except for simple poles at s = 0 and s =1. Supposedly, after accounting for the residue at s = 1, for all $\sigma > 1$, the integral can be moved, from Re(s) = 1/2 to Re(s) = $\sigma$ so that $$Q(x) = -\frac{e^{-1/2x}}{4i}\int_{\sigma - i\infty}^{\sigma + i\infty} \zeta(s)\Gamma(\frac{s}{2})\pi^{-s/2}e^{xs} ds + \frac{e^{-1/2x}}{4i}2\pi iRes[\zeta(s)\Gamma(\frac{s}{2})\pi^{-s/2}e^{xs}].$$

I understand the Residue Theorem, but I still do not understand understand why we can move the integral from Re(s) = 1/2 to Re(s) = $\sigma$.

My guess is that if we consider rectangular contour shown in the picture below, then we want to show that as $T \to \infty$, that the integrals over the horizontal sides of the rectangle approach 0.

If that is the case, then the statement follows from the Residue Theorem. Those latter details I have already worked out. Can anyone help? Is there an upperbound for $\zeta(s)$ and $\Gamma(\frac{s}{2})$ in this region that I am missing?

The classical upper bound $\big|\zeta(\sigma+it)\big|\le O\big(t^{1-\delta}\big)$ for $\sigma\ge\delta$ provides $\big|\zeta(\sigma+it)\big|\le O\big(\sqrt{t}\big)$ which can be sufficient for you. See Titchmarsh's book, formula (9) on page 3.

Instead of the trivial bound $\big|\Gamma(\sigma+it)\big|\le\Gamma(\sigma)$ we need something better, for example, $$\big|\Gamma(s/2)\big| = \left|\frac2s \Gamma(s/2+1)\right| \le \frac2{|t|} \Gamma(\sigma/2+1) \le O(1/t).$$

For real $x$ the other two factors are bounded, so the integrals along the horizontal lines converge to $0$.