Show that $\int_{2}^x\frac{\pi(t)}{t(t-1)}dt=\log \log x+ O(1)$

Question

Do you use the fact that $\pi(t) = \frac{t}{\log t} + O\left(\frac{t}{\log^2t}\right)$ and then

$\int_{2}^x\frac{\pi(t)}{t(t-1)}dt= \int_2^x\left(\frac{t}{\log t} + O\left(\frac{t}{\log^2t}\right)\right)\left(\frac{1}{t(t-1)}\right)dt$

and work from there?

Thank you, would you know how to integrate $\frac{1}{(t-1)\log t}$? and any other methods to find this? — Pablo, Aug 19 '14 at 00:30
$$\frac{1}{(t-1)\log t} = \frac{1}{t\log t} + \frac{1}{t(t-1)\log t}$$ The first is the derivative of $\log \log t$, the second is integrable over $[2,\infty)$. — Daniel Fischer, Aug 19 '14 at 00:33

score 0 · Accepted Answer · answered Aug 19 '14 at 00:34

Since $\frac{1}{t-1}=\frac{1}{t}+O\left(\frac{1}{t^2}\right)$ and $\frac{1}{t^2\log t}$, as well as $\frac{1}{t(t-1)\log t}$, is integrable over $[2,+\infty)$, we have: $$\int_{2}^{x}\frac{dt}{(t-1)\log t}=\int_{2}^{x}\frac{dt}{t\log t}+O(1) = \log\log x + O(1).$$

Show that $\int_{2}^x\frac{\pi(t)}{t(t-1)}dt=\log \log x+ O(1)$

1 Answers1