I'd like to inquire as to the nature of the error term in the Siegel‒Walfisz theorem. I understand that it takes the form $$ O\left( x \exp\left( -C \sqrt{\ln x} \right) \right). $$
Yet this function is not one of those that one studied in one's first term calculus course and knows everything about. (For instance, a version of the prime number theorem gives an error term of $x/(\ln x)^2$, which is a function that seems more familiar.)
My question is thus: How fast does the error term in the Siegel‒Walfisz theorem grow? How does it compare to better known error terms? (I'd be particularly interested in the error term $x/(\ln x)^2$.)
For instance, writing $$ x \exp\left( -C \sqrt{\ln x} \right) = \frac{x}{x^{C/\sqrt{\ln x}}}, $$ we find that it eventually supersedes $x^\alpha$ for any $\alpha < 1$.
To compare to $\frac{x}{(\ln x)^2}$ it is helpful to write $(\ln x)^2$ as an exponential. Namely $(\ln x)^2 = \exp( \ln \ln x)^2 = \exp(2 \ln \ln x) $.
So in the one you divide by $\exp(C \sqrt{\ln x}) $ and in the other by $\exp(2 \ln \ln x)$. Now $\sqrt{\ln x}$ grows much faster than $2 \ln \ln x$, and thus you error-term is smaller (as you divide by something larger).
More generally, the error term is better than $\frac{x}{(\ln x)^k}$ for any positive integer $k$.
So the short is, better than $\frac{x}{(\ln x)^k}$ for any $k \ge 1$ yet worse than $x^{\alpha}$ for $\alpha \lt 1$, as you said.
It is also worse than $x\exp (-C (\ln x)^{\beta}) $ for $1/2 < \beta <1$ (and better for $\beta <1/2$).