Perron's formula where the integrand of the contour is badly behaved at (and left of) zero I am attempting to use Perron's formula to recover the asymptotic form of a summatory function. Namely, it can be shown (is not difficult to prove) that for the prime omega function, $\omega(n)$, its Dirichlet series for $\Re(s) > 1$ is given by 
$$D_{\omega}(s) := \sum_{n \geq 1} \frac{\omega(n)}{n^s} = \zeta(s) P(s),$$ 
where $P(s) := \sum_{p} p^{-s},\ \Re(s) > 1$ is the prime zeta function. For example, this relation can be seen by showing that 
$$\prod_{p\mathrm{\ prime}} \left(1-\frac{u}{1-p^s}\right) = \sum_{n \geq 0} \frac{u^{\omega(n)}}{n^s},$$ and then differentiating with respect to $u$. So, in principle, I should have by Perron's formula that 
$${\sum_{n \leq x}}^{\prime} \omega(n) = \frac{1}{2\pi\imath} \int_{c-\imath\infty}^{c+\imath\infty} D_{\omega}(s) \frac{x^s}{s} ds,$$ 
for suitably large, finite $c > 1$. But now we arrive at a BIG, nay HUGE, complication, which is that due to the nature of its singularities, it is well-known that $P(s)$ cannot be analytically continued at or to the left of zero! I still would like to be able to approximate the contour integral on the right-hand-side of the previous equation. 
The next part of this is my attempt to enable this to happen within some not unreasonable added asymptotic error. Please help me to debug my working lemma to accomplish just this.
There are fairly standard bounds on the prime counting function, $\pi(x)$, for sufficiently large $x \geq 17$: 
$$\frac{x}{\log x} < \pi(x) < C \cdot \frac{x}{\log x}, C \approx 1.25506.$$ 
Now additionally, by a Mellin transform, we can write for all $\Re(s) > 1$ that 
$$P(s) = s \int_1^{\infty} \frac{\pi(x)}{x^{s+1}} dx,$$ 
which is not too bad to evaluate and estimate if we plug in the previous upper and lower bounds for $\pi(x)$. Thus my question (I would love to make a little lemma out of this) is the following: 
Proposed Lemma: Suppose that $$|R_1(s)| < |P(s)| < |R_2(s)|,$$ for all $\Re(s) > 1$, and moreover, the functions $R_1(s),R_2(s)$ can both be analytically continued to the entire complex plane, with the exception of at finitely many poles where we consider these functions to be undefined. Then for large enough (but finite) real $c > \sigma_P$, do I obtain that the contour integrals are bounded as follows:
$$\left\lvert \frac{1}{2\pi\imath} \int_{c-\imath\infty}^{c+\imath\infty} R_1(s) \zeta(s) \frac{x^s}{s} ds\right\rvert < {\sum_{n \leq x}}^{\prime} \omega(n) < \left\lvert \frac{1}{2\pi\imath} \int_{c-\imath\infty}^{c+\imath\infty} R_2(s) \zeta(s) \frac{x^s}{s} ds\right\rvert.$$ 
Are there any additional necessary conditions that need to be placed on the functions $R_1(s),R_2(s)$ to give truth to the previous inequalities? 
Thanks in advance. I really do have a good application in mind for this lemma.
 A: Let $C_T= \{ \sigma+it,t \in [-T,T], \sigma = 1-\frac{A}{\log(3+|t|)}\}$ where $A$ is found from the zero-free region detailed in the proof of the PNT.
$L(s) =\sum_{n=2}^\infty \frac{n^{-s}}{\log n}$ then $L(s)-\log(s-1),\log ((s-1) \zeta(s))$, $\log \zeta(s)-P(s)$ and $\zeta(s)P(s)-\zeta(s)L(s)=\sum_{n=1}^\infty a_n n^{-s}$ are analytic on some right plane containing $C_\infty$, thus
$$\sum_{n \le x}\omega(n) = \frac{1}{2i \pi} \int_{2-i\infty}^{2+i\infty} D_{\omega}(s) \frac{x^s}{s} ds $$
$$ = \frac{1}{2i \pi} \int_{C_\infty} (D_{\omega}(s)-\zeta(s) L(s)) \frac{x^s}{s} ds + \frac{1}{2i \pi} \int_{2-i\infty}^{2+i\infty} \zeta(s) L(s) \frac{x^s}{s} ds $$
$$ =  \frac{1}{2i \pi} \int_{C_T} (D_{\omega}(s)-\zeta(s) L(s)) \frac{x^s}{s} ds+O(\int_{1-\frac{A}{\log(3+|T|)}+iT}^{r+iT}|(D_{\omega}(s)-\zeta(s) L(s)) \frac{x^s}{s} ds|) \\+ \sum_{n=1}^\infty O(a_n \int_{d+iT}^{d+i \infty} \frac{(x/n)^s}{s}ds)+
 \sum_{n \le x} a_n$$
each term can be estimated to obtain 
$$\sum_{n \le x}\omega(n) =  \sum_{n \le x} a_n + O(\frac{x}{\log^k x}), \qquad\forall k$$
