Simplifying a log log function This starts with the summation function over $\omega(n)$:
$$
S_\omega(x) = \sum_{n = 1}^{x} \omega(n) 
  = x \log \log x + B_1 x + o(x)
$$
I was trying to separate this into congruence classes modulo 6. (If there's a straight-forward way, let me know and I'll ask that question too.)
Now I am focused on the following, and have made progress (note the 6x limit):
$$
S_\omega(x) = S_\hat{0}(x) + S_\hat{1}(x) \; + \;... + \; S_\hat{5}(x) \\
S_\chi(6x) = \sum_{n = 1}^{6x}{\omega(n)\chi_1} = S_\hat{1}(6x) + S_\hat{5}(6x) \\
 (several \; skipped \; steps \; here) \\
S_\chi(6x) = S_\omega(6x) + S_\omega(x) - S_\omega(3x)
  -  S_\omega(2x) - \left\lfloor\frac{5x}{3} \right\rfloor
$$
That is as clean as I can get it, so I substitute the approximations in:
$$
S_\chi(6x) = 6x \log\log 6x + x \log\log x - 3x \log\log 3x - 2x \log\log 2x \\
+ B_1 2x - \left\lfloor\frac{5x}{3}\right\rfloor + o(x)
$$
Everything I try from here either breaks it, or makes is worse. Can this be simplified?
 A: If you're happy with an $o(x)$ error term then note that
$$
   x \log \log x - x \log \log cx = x \log \frac{\log x}{\log cx} = x \log\left(1 - \frac{\log c}{\log x + \log c}\right).
$$
In the limit as $t \to \infty$, we have $\log (1 + t) \sim t$, so this is asymptotically $-\frac{x \log c}{\log x + \log c} = O(\frac{x}{\log x})$.
Therefore we can ignore the difference between $\log \log x$, $\log \log 6x$, and so forth. Instead of having $x\log \log x$, $2x\log \log 2x$, $3x\log\log3x$, and $6x\log\log6x$, we just have $x\log\log x$, $2x\log\log x$, $3x\log\log x$, and $6x\log\log x$, which are now like terms that cancel. Continuing from the last line in your question, we get
\begin{align}
S_\chi(6x) &= 6x \log\log 6x + x \log\log x - 3x \log\log 3x - 2x \log\log 2x \\
& \qquad + B_1 2x - \left\lfloor\tfrac{5x}{3}\right\rfloor + o(x) \\
  &= 6 x \log \log \color{red}{x} + x \log \log x - 3x\log\log\color{red}{x} - 2x \log\log\color{red}{x} \\
& \qquad + B_1 2x - \tfrac53x + o(x) \\
 &= (6 + 1 - 3 - 2) x \log \log x + B_1 2x - \tfrac53x + o(x)
\end{align}
giving us the final answer of $S_\chi(6x) = 2x\log\log x + (2B_1 - \frac53)x + o(x)$.
A: I did not find a way to simplify the approximation enough to be usable. Fortunately I can use an upper or lower bound. 
As a summation function involving positive integers, $S_\omega(x)$ is strictly increasing. And from the approximation, $S_\omega'(x)$ is also increasing.
$$
S_\omega(x) > S_\omega\!\left(\frac{x}{n}\right) n \\
S_\omega(x) - S_\omega\!\left(\frac{x}{n}\right) > S_\omega\!\left(x - \frac{x}{n}\right)
$$
Applying this to the above:
$$
S_\chi(x) = S_\omega(x) - S_\omega\!\left(\frac{x}{2}\right) - S_\omega\!\left(\frac{x}{3}\right)
   + S_\omega\!\left(\frac{x}{6}\right) - \frac{5x}{18} \\
S_\chi(x) > \quad S_\omega\!\left(\frac{x}{2}\right) \quad - \quad S_\omega\!\left(\frac{x}{6}\right) \quad - \frac{5x}{18} \\
S_\chi(x) > \quad S_\omega\!\left(\frac{x}{3}\right) \quad - \frac{5x}{18} \\
S_\chi(x) > \frac{x}{3} \left( \log \log \frac{x}{3} + B_1 - \frac{5}{6} \right) + o(x)
$$
This is a rather loose lower bounds and is a "NO" answer to the original question so I'll wait before accepting it. 
