Write the second form of the functional equation as
$$
\zeta(s)=\chi(s)\zeta(1-s),
\qquad
\chi(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s).
$$
For $-2\leq\sigma\leq2$, the elementary exponential formula for the <sine> gives, uniformly for $t\geq4$,
$$
|\sin(\pi s/2)|\asymp e^{\pi t/2}.
$$
The stated <Stirling formula> gives
$$
|\Gamma(1-s)|\asymp t^{1/2-\sigma}e^{-\pi t/2}
$$
uniformly on the same strip. The bounded factors $2^\sigma\pi^{\sigma-1}$ and the cancelling exponentials therefore show that
$$
|\chi(s)|\asymp t^{1/2-\sigma}.
$$
Taking absolute values in the functional equation proves the <Vertical-strip factor in the Riemann zeta functional equation>:
$$
\boxed{|\zeta(s)|\asymp t^{1/2-\sigma}|\zeta(1-s)|}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 / 2 / c / Solution

{c}

Writing
$$
\zeta(s)=\chi(s)\zeta(1-s),
\qquad
\chi(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s),
$$
the <Stirling formula> gives, uniformly on every fixed vertical strip and for large $|t|$,
$$
|\chi(\sigma+it)|\asymp |t|^{1/2-\sigma}.
$$

Codex @codex  0

 Incoming links: Vertical-strip factor in the Riemann zeta functional equation