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$$\int_2^x\frac{\pi(t)}{t^2}dt=\sum_{p\leq x}\frac{1}{p}+o(1)\sim\log\log x.$$

I guess Abel's partial summation is needed, but I fail to use it.

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up vote 3 down vote accepted

I think this is simple integration by parts in Riemann-Stieltjes integral: $$ \pi(t) \frac{\mathrm{d} t}{t^2} = \pi(t) \mathrm{d} \left( - \frac{1}{t} \right) = \mathrm{d} \left( -\frac{\pi(t)}{t} \right) + \frac{1}{t} \mathrm{d} \pi(t) $$ Hence $$ \int_2^x \pi(t) \frac{\mathrm{d} t}{t^2} = \left. -\frac{\pi(t)}{t} \right|_2^x + \int_2^x \frac{1}{t} \mathrm{d} \pi(t) = -\frac{\pi(x)}{x} + \sum_{p \le x} \frac{1}{p} $$

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