Doubt in a lemma which is to be used in deriving Perron's Formula from Tom M Apostol While studying analytic number theory from Tom M Apostol I could think about inequalities which are deduced by Apostol in lemma 4 of chapter 11 . 
The lemma and doubts in proof are->



My doubt is - Just taking 1 st integral on RHS ( in image 2 ) I have following doubts ->
  
  
*
  
*I know T is constant in 1st integral in limit but how can it be removed from limit, knowing that it tends to $\infty$? 
  
*mainly I m unable to understand how to derive 
  1st line of  inequalities on Page 2 . 
  

Can someone please explain!! 
 A: *

*$T$ isn't being removed. Nor is it tending to anything. Only three facts are being used in going from the line above "Let $b\to\infty$": firstly, that $\log(a) = -\log(1/a)$ (to get rid of the minus sign on the top of the first fraction), secondly, that $\frac{2Ta^b}{b}\to 0$ as $b\to \infty$ (to get rid of the second fraction), and thirdly the fact that we can pass from $A \leq B(b)$ for all $b$ to $A \leq \lim_{b\to\infty}B(b)$. 

*To derive the initial inequality, we have 
\begin{align*}\left|\int_{c-iT}^{c+iT}a^z\frac{dz}{z}\right| &= \left|\int_{b+iT}^{c+iT}a^z\frac{dz}{z} + \int_{b-iT}^{b+iT}a^z\frac{dz}{z} + \int_{c-iT}^{b-iT}a^z\frac{dz}{z}\right|\\
&\leq \left|\int_{b+iT}^{c+iT}a^z\frac{dz}{z}\right|+\left|\int_{b-iT}^{b+iT}a^z\frac{dz}{z}\right|+\left|\int_{c-iT}^{b-iT}a^z\frac{dz}{z}\right|\\
\end{align*}
So now we just need to show that the middle term is bounded above by $\displaystyle\frac{2Ta^b}{b}$, while the other two are each bounded above by $$\int_{b}^c\frac{a^x}{T}dx.$$
We start with the former: \begin{align*}\left|\int_{b-iT}^{b+iT}a^z\frac{dz}{z}\right| &= \left|\int_{-T}^T\frac{a^{b+ix}}{b+ix}dx\right|\\&\leq \int_{-T}^T\left|\frac{a^{b+ix}}{b+ix}\right|dx\\&= \int_{-T}^T\frac{a^b}{|b+ix|}dx\\&\leq \int_{-T}^T\frac{a^b}{b}dx\\&=\frac{2Ta^b}{b}.\end{align*}
Now, the first and last summand:
\begin{align*}
\left|\int_{b\pm iT}^{c\pm iT}a^z\frac{dz}{z}\right| &= \left|\int_{b}^c\frac{a^{x\pm iT}}{x\pm iT}dx\right|\\&\leq \int_b^c\left|\frac{a^{x\pm iT}}{x\pm iT}\right|dx\\
&=\int_b^c\frac{a^x}{|x\pm iT|}dx\\
&= \int_{b}^c\frac{a^x}{T}dx,
\end{align*}
as required. 
