Show $O\left(\sqrt x\right) + \int_{\sqrt{x}}^{x}\frac{\pi(t)}{t}dt <<\frac{x}{\ln x}$

Question

Show the following:$$ O\left(\sqrt x\right) + \int_{\sqrt{x}}^{x}\frac{\pi(t)}{t}dt <<\frac{x}{\ln x}$$

where $\pi{(x)}$ prime counting function. This is the last step in a derivation and I'm stuck on it.

Answer 1

Want $O\left(\sqrt x\right) + \int_{\sqrt{x}}^{x}\frac{\pi(t)}{t}dt <<\frac{x}{\ln x} $.

Since $\sqrt{x} =o(x/\ln(x)) $, that part doesn't matter. And, actually, $\sqrt{x} =o(x^{c+1/2}/\ln(x)) $ for any $c > 0$.

As for the $\int_{\sqrt{x}}^{x}\frac{\pi(t)}{t}dt $ term, Chebychev showed by elementary (non-complex) means that $\dfrac78 \lt \dfrac{\pi(n)}{n/\ln(n)} \lt \dfrac98 $, so that $\int_{\sqrt{x}}^{x}\frac{\pi(t)}{t}dt \lt \int_{\sqrt{x}}^{x}\frac98\frac{t/\ln(t)}{t}dt = \frac98\int_{\sqrt{x}}^{x}\frac{1}{\ln t}dt $ and $\int_{\sqrt{x}}^{x}\frac{\pi(t)}{t}dt \gt \frac78\int_{\sqrt{x}}^{x}\frac{1}{\ln t}dt $.

Since $\int_c^x \dfrac{dt}{\ln(t)} \approx \dfrac{x}{\ln(x)}+\dfrac{2x}{\ln^2(x)}+O(\dfrac{x}{\ln^3(x)}) $, $\int_{\sqrt{x}}^x \dfrac{dt}{\ln(t)} \approx \dfrac{x}{\ln(x)}+o(\dfrac{x}{\ln(x)}) $, so I don't see where the $<<$ comes from.