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The Chebyshev functions are defined as $\psi(x) = \sum_{p^m \leq x} \log n$ and $\theta(x) = \sum_{p\leq x} \log p$, where $p$ is a prime, $m\geq 1$ is an integer and $n=p^m$ in $\psi(x)$. It is known that there exist positive constants $c_{1}$ and $c_2$ such that $c_{1}x < \psi(x) < c_{2} x$ and $ \frac{1}{2}c_{1}x < \theta(x) < c_{2} x$. By these bounds we find that $ \dfrac{ \psi(x) - x}{\psi(x) - \theta(x)} < \dfrac{(c_{2} -1)x}{(c_2 - c_1)x} = \dfrac{c_{2} -1 }{c_2 - c_1} <\infty $.

Hence invoking the well known result $\mid \psi(x) - \theta(x) \mid = O(x^{\frac{1}{2}}\log x)$, it then follows that $\mid \psi(x) - x \mid = O(x^{\frac{1}{2}}\log x)$ ?

EDIT: It is also known that $\theta (x)$ tends to $x$ as $x$ tends to $\infty$. Surely, an impication of this is $\dfrac{ \psi(x) - x}{\psi(x) - \theta(x)} < \infty$ since $\psi(x) \neq \theta(x)$.

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  • $\begingroup$ Note that $c_1 - c_2 < 0$. You don't have a lower bound $\psi(x) - \vartheta(x) > \gamma\cdot x$. $\endgroup$ – Daniel Fischer Feb 6 '16 at 20:33
  • $\begingroup$ I think you're referring to the typo i just edited. $\endgroup$ – User1 Feb 6 '16 at 20:37
  • $\begingroup$ In that case, what's your argument for $$\frac{\psi(x) - x}{\psi(x) - \vartheta(x)} < \frac{(c_2-1)x}{(c_2-c_1)x}\,?$$ $\endgroup$ – Daniel Fischer Feb 6 '16 at 20:39
  • $\begingroup$ The $x's$ on the RHS cancel and we remain with a constant. $\endgroup$ – User1 Feb 6 '16 at 20:45
  • $\begingroup$ You do that after the step I'm asking about. One thing after the other. $\endgroup$ – Daniel Fischer Feb 6 '16 at 20:46
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The inequality \[\frac{\psi(x) - x}{\psi(x) - \vartheta(x)} < \frac{(c_2 - 1)x}{(c_2 - c_1)x}\] is false, as Daniel Fischer mentions. The mistake is in the denominator; you know that $c_1 x < \psi(x) < c_2 x$ and $\frac{1}{2} c_1 x < \vartheta(x) < c_2 x$, but this does not imply that \[\frac{1}{\psi(x) - \vartheta(x)} < \frac{1}{(c_2 - c_1)x},\] or equivalently that \[\psi(x) - \vartheta(x) > (c_2 - c_1)x.\] In fact, \[\psi(x) - \vartheta(x) = \sum_{\substack{p^m \leq x \\ m \geq 2}} \log p,\] and from this it is easy to show that \[\frac{1}{2} c_1 \sqrt{x} < \psi(x) - \vartheta(x) < c_3 \sqrt{x} \log x,\] so the inequality is far from true.

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