Perron's Formula (Segal, pg 622): Let $w=u+iv $ , $f(w) = \sum_{n=1}^\infty \frac{a_n}{n^w}$ for $u > 1$, where $a_n = O (A(n))$ as $N \rightarrow \infty$, and $A(n)$ is non-decreasing; and for some $\alpha> 0$, $$ \sum_{n=1}^\infty \frac{|a_n|}{n^u} = O\Big( \frac{1}{(u-1)^\alpha} \Big) $$ as $u \rightarrow 1^{+}.$ Then, if $c > 0$, $\sigma+c > 1$, $x$ is not an integer, $N$ is the nearest integer to $x$, and $T>0$, $$ \sum_{n < x } \frac{a_n}{n^s} = \frac{1}{2 \pi i } \int_{c-iT}^{c+iT} f(s+w) \frac{x^w}{w} \, dw + O\Big( \frac{x^c}{T(\sigma+c-1)^{\alpha}} \Big) + O\Big( \frac{A(2x)x^{1 - \sigma} \log x }{T} \Big) + O \Big( \frac{A(N) x^{1- \sigma} }{T|x-N|} \Big) $$ where $c$ may depend upon $x$.
I am confused of the following points in this Theorem:
What do the Big-$O$s mean in the last equation? Are we limiting both $x \rightarrow +\infty$ and $ \sigma + c \rightarrow 1^{+}$?
Why does $c$ depend on $x$?
I searched on Perron's Formula which seems different (?); also the theorem from segal doesn't consider the case when $x$ is integer.
The usual Perron formula is just the inverse Mellin/Laplace/Fourier transform applied to $\frac{f(\sigma+it)}{\sigma+it}$ the Fourier transform of $\sum_{n < e^u} a_n e^{-\sigma u}$
The zero-free region for $\zeta(s),s = \sigma+it$ is $\sigma > 1-\frac{1}{1+A \log |t|}$
This is an improved Perron formula (with all the $O$ terms) allowing us (with $f(s) = \frac{\zeta'(s)}{\zeta(s)}$ or $\frac{\zeta'(s)}{\zeta(s)}+ \frac{1}{s-1}$ or $\frac{1}{\zeta(s)}$) to derive the prime number theorem from : the zero-free region and some bounds $\zeta'(s) = \mathcal{O}(\log |\Im(s)|^k),\frac{1}{\zeta(s)} = \mathcal{O}(\log |\Im(s)|^k)$ in this region.
$c$ depends on $x$ because when $x$ is large we need $T$ large, unfortunately our zero-free region depends on $\sigma$ and requires $c < A/ \log |T|$
Here the $O$ constants should not depend on $x,T$, not sure for $c$, see the derivation in your book