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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

Alain Remillard
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idontknowanything
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    Why is it not just zero since the set of primes is discrete? Or do you really intend for it to be a sum? Or is the integral really going up to $x$ and $p_u$ is actually $p_{\lfloor u \rfloor}$ or something like this? – Ian Feb 18 '21 at 19:13

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Hint

for $u\ge 6$

$$u (\log (u)+\log (\log (u))-1)\le p_u\le u (\log (u)+\log (\log (u)))$$

thus $$\frac{u(1-\log p_u)}{p_u^2}\sim -\frac{1}{u\log u}$$

see here. First answer.

Raffaele
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