Take $1\leq H\leq X$. Let $f(u)$ be a smooth function on $(X,2X)$ and satisfy there $f^{(j)}(u)\ll _jH^j/X^j$ for all $j$. Let $$\mathfrak g(s)=\frac {\Gamma (s/2)^3}{\Gamma ((1-s)/2)^3}\approx t^{3(\sigma -1/2)}$$ and for $N>0$ $$\mathcal M(s)=\int _0^\infty f(u)u^sdu$$ and $$\mathcal I(N)=\int _{(1/8)} \frac {\mathfrak g(s)f(-s)ds}{N^s}.$$
On page 2 at the end of Section 2 in https://arxiv.org/pdf/1304.0377.pdf the author says that $\mathcal I(N)$ is negligibly small (I assume, smaller than any power of $X$ say) once $N$ is larger than essentially $H/X$. Why is this true?
I can see it's true for $N$ larger than $2H^3/X$ by this argument: Choose $\sigma $ so large that say $2^\sigma >X^{2023}$ and choose $j$ so large that $\delta :=j-3(\sigma -1/2)$ is a bit bigger than 1. Then \begin{eqnarray*} \mathcal I(N)&\approx &\int _{(\epsilon )}t^{3(\sigma -1/2)}N^{-\sigma }\int _0^\infty f(u)u^{-s}duds \\\ &=&\int _{(\epsilon )}t^{3(\sigma -1/2)}N^{-\sigma }\int _0^\infty \frac {f^{(j)}(u)u^{-s+j}duds}{(-s+1)\cdot \cdot \cdot (-s+j)} \\\ &\ll &X(X/H)^{-j}\int _{(\epsilon )}t^{3(\sigma -1/2)-j}N^{-\sigma }X^{-\sigma +j}dx \\\ &\ll &XH^j(NX)^{-\sigma } \\\ &=&XH^{\delta -3/2}\left (\frac {H^3}{NX}\right )^\sigma \\\ &\ll &\frac {X}{2^{\sigma }}\ll \frac {1}{X^{2023}} \end{eqnarray*}