I've been self studying the Montgomery-Vaughan book and after much thought I've come to a dead end with the part of the proof of the Theorem $5.2$. The theorem says:
Theorem 5.2 : If $σ_0>\max(0,σ_a)$ and $x>0$, then$$\sum_{n≤x}'a_n=\frac{1}{2πi}\int_{σ0+iT}^{σ0−iT}α(s)\frac{x^s}{s}ds+R$$ where$$R=\frac{1}{π}\sum_{x/2\leq n\leq x}a_n \operatorname{si}(T\log\frac{x}{n})−\frac{1}{π}\sum_{x\leq n \leq 2x}a_n \operatorname{si}(T\log\frac{n}{x})+ \mathcal{O}(\frac{4^{σ_0}+x^{σ_0}}{T}\sum_{n}\frac{|a_n|}{n^{σ_0}}).$$
They proceed with the proof as follows:
Proof: Since the series $α(s)= \sum_{n} \frac{a_n}{n^s}$ is absolutely convergent on the interval $ [σ_0−iT,σ_0+iT]$, we see that $$\frac{1}{2πi}\int_{σ_0+iT}^{σ_0−iT} α(s)\frac{x^s}{s}ds=\sum_{n}a_n \frac{1}{2πi}\int_{σ_0+iT}^{σ_0−iT}(\frac{x}{n})^s\frac{ds}{s}. $$ Thus it suffices to show that \begin{equation} \frac{1}{2πi}\int_{σ_0+iT}^{σ_0−iT} \frac{y^s}{s}ds = \begin{cases} 1+\mathcal{O}(\frac{y^{σ_0}}{T}) & y \geq 2 \\ 1+\frac{1}{π}\operatorname{si}(T\log y)+\mathcal{O}(\frac{2^{σ_0}}{T}) & 1\leq y \leq 2 \\ \frac{−1}{π} \operatorname{si}(T\log \frac{1}{y})+\mathcal{O}(\frac{2^{σ_0}}{T})& 1/2 \leq y \leq 1 \\ \mathcal{O}(\frac{y^{σ_0}}{T})& y \leq 1/2 \end{cases} \end{equation} for $σ_0>0$.
All is fine until for the second case he starts:
Suppose now that $1\leq y \leq 2$, and take $C$ to be the closed rectangular path from$ σ_0−iT$ to $σ_0+iT$ to $iT$ to $−iT$ to $σ_0−iT$, with a semicircular indentation of radius $\epsilon$ at $s=0$. Then by Cauchy’s theorem $$\frac{1}{2πi}\int_{C}\frac{y^s}{s}ds =0.$$ We note that $$ \int_{σ_0±iT}^{±iT}\frac{y^s}{s}ds \ll \frac{1}{T} \int_{0}^{σ_0} y^σ dσ \leq \frac{1}{T} \int_{0}^{σ_0}2^σ dσ = \frac{2^{σ_0}}{T}.$$ The integral around the semicircle tends to $1/2$ as $\epsilon \to 0$, and the remaining integral is \begin{equation} \label{eq1} \begin{split} \frac{1}{2πi} \lim_{\epsilon \to 0} \int_{i\epsilon}^{iT}+\int_{-iT}^{−i\epsilon} \frac{y^s}{s}ds & = \frac{1}{2πi} \lim_{\epsilon \to 0} \int_{\epsilon}^{T} (y^{it}−y^{−it})\frac{dt}{t} \\ & = \frac{1}{π} \int_{0}^{T\log(y)} \sin{v} \frac{dv}{v} \\ & = \frac{1}{2}+ \frac{1}{π} \operatorname{si}(T\log{y}) \end{split} \end{equation} by $\operatorname{si}(x)+\operatorname{si}(−x)=−\int_{-\infty}^{\infty}\frac{\sin{u}}{u}du=−π$. This gives the value we need when $1 \leq y \leq 2$ and the case $1/2 \leq y \leq 1$ is treated similarly.
I have two question which I have not managed to get my head around more specifically my problem starts where they start computing the 'remaining integrals'. I would appreciate if anyone could give some detailed explanation of what the proof is doing? my next question is whether for the case $1/2 \leq y \leq 1$ the contour we take into account is the same as the contour $C$?
Please note that $$\operatorname{si}(x)=−\int_{x}^{\infty}\frac{\sin{u}}{u}du.$$
Thank you so much in advance.
The motivation is that when $y>1$ the function $y^s/s$ will go to zero when $\Re(s)\to+\infty%$ and when $y<1$ the function vanishes when $\Re(s)\to-\infty$. This indicates that different rectangular paths need to be constructed so that we get into the following type of situation: $$ \oint=\int_{\gamma_1}+\int_{\gamma_2}, $$ where $\gamma_1$ denotes the original path of integration, $\gamma_2$ is constructed so that $\gamma_1+\gamma_2$ becomes a closed path. This means we can evaluate the left hand side using residue theorem to get the main term in the asymptotic expansion of integral over $\gamma_1$. As a result, the integral ("the remaining integral") over $\gamma_2$ is believed to be the error term.