I have two questions regarding the proof of Theorem 4.15. of Titchmarsh's book The Theory of the Riemann Zeta-Function:
Question 1: Integral on the counter $C$ in the line 7th of the picture [The first green-underlined] is equal to the sum of integral on the new counter (which is the four straight lines) plus $2 \pi i$ times residues of $\dfrac{w^{s-1}e^{-mw}}{e^w-1}$. Considering that the domain between the two counters consists of simple poles at $\pm 2 \pi i n$, $n=1, 2, \dots, q$, so I got for the second term of RHS of $\zeta(s)$ in the line 16th [The second green-underlined]: $$\bigg( \sum_{n=1}^q (2 \pi i n)^{s-1} + \sum_{n=1}^q (-2 \pi i n)^{s-1} \bigg) e^{- \pi i s} \Gamma(1-s) = \dfrac{1}{2} e^{- \pi i s} \chi(s) \sum_{n=1}^q \dfrac{1}{n^{1-s}},$$ where $\chi(s)=2^s {\pi}^{s-1} \sin \frac12 \pi s \Gamma(1-s).$ I did calculations twice and why am I getting a different result?
Question 2: In the first orange-underlined equation the term ${\eta}^{\sigma -1} e^{\frac{\lambda t}{\eta \sqrt{2}}}$ appears and in the second orange-underlined equation it disappears. How $\dfrac{e^{-mw}}{e^w-1} = O(e^{\frac{-\lambda t}{\eta \sqrt{2}}})$ makes that term to vanish? Firstly, sum of two $O$ terms may not cancel even if they are negative of each other; secondly, the first one has ${\eta}^{\sigma -1}$ as coefficient but the second one doesn't.
For question 1, notice that
\begin{aligned} (2\pi i)^{s-1}+(2\pi i^{-1})^{s-1} &=(2\pi)^{s-1}(i^{s-1}+i^{1-s})=(2\pi)^{s-1}(i^s-i^{-s})i^{-1} \\ &=2^s\pi^{s-1}{e^{\pi is/2}-e^{-\pi is/2}\over2i}=2^s\pi^{s-1}\sin\left(\pi s\over2\right). \end{aligned}
For question 2, the O-terms are multiplied instead of added. In general, when $f$ and $g$ are real functions, it is always true that $O(e^f)\cdot O(e^g)=O(e^{f+g})$.