For $x>0$ and $0<\sigma <1/6$ consider $$\int _{\sigma \pm i\infty }\underbrace {\frac {\Gamma ^3(s/2)}{\Gamma ^3((1-s)/2)}}_{=:G}\frac {ds}{x^s}$$ which is absolutely convergent since $G\asymp |t|^{3\sigma -3/2}$ for large $|t|$.
Let $\epsilon >0$ be given. In https://www.cambridge.org/core/books/abs/sieve-methods-exponential-sums-and-their-applications-in-number-theory/on-the-ternary-additive-divisor-problem-and-the-sixth-moment-of-the-zetafunction/DBEDC56531F9A6C7466D0018DB10CC31 the author states (equation (3.17) and definition (3.7)) that with a trivial estimation the above integral is $\ll x^\epsilon $. My question is why is this true?
What I know/understand: For $|t|\geq 1$ we have the above bound for $G$, whilst for $|t|\leq 1$ we have $G\asymp 1/|s|^3$ so if $x\gg 1$ then we can just choose say $\epsilon =1/10$ to see that the integral is $$\ll x^{-1/10}\int _{1/10\pm i}\frac {ds}{|s|^3}+\int _{1/10+i}^{1/10+i\infty }t^{3/10-3/2}dt\ll 1\ll x^\epsilon $$ so I'm ok in that case. My issue is with small $x$, where the claimed bound is stronger than $\ll 1$. (Also - the author really does wants arbitrarily small $x$, so it's not "he probably means $x>x_0$" or something).
I think for small $x$ I could choose $\epsilon =1/|\log x|$ so that $1/x^\epsilon =e$ so that the part of the integral with $|t|\geq 1$ is again $\ll 1$ leaving me with just the part $$\int _{1/\log x\pm i}\frac {Gds}{x^s}$$ left to bound. But now I'm obviously doing something wrong because if I deform this path into a half-circle to the left of zero, then the integrand will be $\ll 1/x^{-1}+1/x^\epsilon \ll 1$ which is fine, but I'd've picked a residue at $s=0$ which gives the true size of the integral to be $Res_{s=0}G/x^s\approx (\log x)^2$, which contradicts the claim. (I also think the author would have mentioned the argument if it involved deforming the integral, so I'm assuming there's something easier I'm missing.)
What am I not doing correctly or misunderstanding?
The trivial estimate is
$$|\int _{\Re(s)=\sigma }\frac {\Gamma ^3(s/2)}{\Gamma ^3((1-s)/2)}\frac {ds}{x^s}|\le x^{-\sigma}\int_{\Re(s)=\sigma }|\frac {\Gamma ^3(s/2)}{\Gamma ^3((1-s)/2)}| d|s| =O(x^{-\sigma})$$ To improve it as $x\to 0$ shift the integral to the left $$\int_{\Re(s)=\sigma }\frac {\Gamma ^3(s/2)}{\Gamma ^3((1-s)/2)}\frac {ds}{x^s}$$ $$=\int_{\Re(s)=-1/2}\frac {\Gamma ^3(s/2)}{\Gamma ^3((1-s)/2)}\frac {ds}{x^s} + 2i\pi Res(\frac {\Gamma ^3(s/2)}{\Gamma ^3((1-s)/2)}x^{-s},0)$$ $$ = O(x^{1/2})+4 \pi^{-3/2}\log^2 x+O(\log x)$$ in particular it is not true that the integral is $O(x^\epsilon)$ as $x\to 0$.