I have some trouble with paper I'm reading.
The goal is this: let $s=\frac{1}{2}+\frac{1}{\log n}+it$.
$M$ is a function such that $M(s)=O(\log^{3}(N(|t|+2)))$.
Define $$U(s)=\frac{1}{2\pi i}\int^{\infty}_{t=-\infty}\left(\frac{1}{N}+2\pi \alpha i\right)^{-s}\Gamma(s)M(s)ds.$$
Then, $$\int^{\frac{1}{q\tau}}_{-\frac{1}{q\tau}}|U(s)|^{2}ds \ll \frac{N}{q\tau}\log^{13}N.$$
Also, $$q\le \tau,\qquad \log^{10}N\le \tau \le N^{1/6}.$$
The paper said that $$U(s)=\frac{1}{2\pi i}\int^{N_0}_{t=-N_1}\left(\frac{1}{N}+2\pi \alpha i\right)^{-s}\Gamma(s)M(s)ds+O(e^{-log^{2}N}),$$ but I can't understand it.
Can you help me?
I'm not sure the meanings of $N_0$ or $n_1$. I only get the error term can be $O(e^{-cN_0}\log^3N)$ for some constant $c>0$.Using the asymptotic formula toward $\Gamma(s)$, we can derive that the upper integral $\int_{N_0}$ is bounded by $\ll \int_{N_0}\log^3N(|t|+2)e^{-t(\pi/2-\arctan 2\pi \alpha n)+\log t/\log n-(1/2+1/\log n)\sqrt{1/n^2+4\pi^2 \alpha^2}} $. If no phase exists ,(ie. may suppose $N_0\gg 1/\log n$, $n>1$) the bound $O(e^{-(\pi/2-\arctan 2\pi \alpha n)/100 \cdot N_0}\log^3N)$ for this integral looks obvious.