Does the following integral converge? If so, is it nonzero? What can we say about the integral as $x\to \infty?$
$$\int_{I} \frac{u-u\log{p_u}}{(p_u)^2}du$$ where $I= \{p \leq x | p\text{ prime}\}$ for some $x$.
Here $p_u$ denotes the $u$:th prime
Does the following integral converge? If so, is it nonzero? What can we say about the integral as $x\to \infty?$
$$\int_{I} \frac{u-u\log{p_u}}{(p_u)^2}du$$ where $I= \{p \leq x | p\text{ prime}\}$ for some $x$.
Here $p_u$ denotes the $u$:th prime
Hint
for $u\ge 6$
$$u (\log (u)+\log (\log (u))-1)\le p_u\le u (\log (u)+\log (\log (u)))$$
thus $$\frac{u(1-\log p_u)}{p_u^2}\sim -\frac{1}{u\log u}$$