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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. enter image description here

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?

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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 $0$.

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