# Bound integral, Riemann hypothesis and Big O notation

Exercise. If Riemann hypothesis holds then we can prove$$\theta(x):=\sum_{p\leq x}\log p = x + \mathcal{O}(x^{1/2}\log^2x).$$ Admitting this result prove that if the Riemann hypothesis is true, then there is a constant A such that $$\sum_{p\leq x}\frac{\log p}{p} = \log x + A + \mathcal{O}(x^{-1/2}\log^2 x)$$

Solution. Let $a_n = I_{\mathbb{P}}(n)\log n$, where $I_{\mathbb{P}}(n)$ prime indicator fucntion. Using Abel's summation formula (taking $f(n)=\dfrac{1}{n}$) we get

\begin{align*} \sum_{p\leq x}\frac{\log p}{p} &= \sum_{n\leq x}\frac{a_n}{n} = \frac{\theta(x)}{x} - \int_1^x \frac{-\theta(t)}{t^2}dt =\\ &= \frac{x + \mathcal{O}(x^{1/2}\log^2x)}{x} + \int_1^x \frac{t + \mathcal{O}(t^{1/2}\log^2t)}{t^2}dt\\ &= 1+\mathcal{O}(x^{-1/2}\log^2 x) + \int_1^x \frac{1}{t} + \frac{\mathcal{O}(t^{1/2}\log^2 t)}{t^2}dt =\\ &=\log x + 1 + \mathcal{O}(x^{-1/2}\log^2 t) + \int_1^x \frac{\mathcal{O}(t^{1/2}\log^2 t)}{t^2}dt \end{align*} My problem starts now. I know $\mathcal{O}(t^{1/2}\log^2 t)$ is map $U(t)$ which verifies $\exists t_0 \in R, C\in R_{>0}$ such than $|U(t)|\leq Ct^{1/2}\log^2 t$ $\forall t>t_0$. In fact, $U(t)=\theta(t)-t$. If I try to integrate directly in Big-O notation I get $\mathcal{O}(1)$, so I think I should bound this \begin{gather*} \int_1^x \frac{\mathcal{O}(t^{1/2}\log^2 t)}{t^2}dt = \int_1^x \frac{U(t)}{t^2}dt = \int_1^\infty \frac{U(t)}{t^2}dt - \int_x^\infty \frac{U(t)}{t^2}dt \end{gather*} But I don't know how.

$$\sum_{p \le x} \frac{\log p}{p} = \int_1^x \frac{1}{t} d \theta(t)= \frac{1}{t} \theta(t)-\int_1^x \theta(t)d\frac{1}{t}$$ $$= 1+\mathcal{O}(x^{-1/2}\log^2 x )+\int_1^x (t+\mathcal{O}(t^{1/2} \log^2 t))\frac{1}{t^2} dt$$ $$= \log x+C+\mathcal{O}(x^{-1/2}\log^2 x ), \qquad C = 1+\int_1^\infty (\theta(t)-t)\frac{1}{t^2} dt$$
• It's remain to prove $\int_1^\infty U(t)/t^2 dt$ is convergent, no? – Rafael Gonzalez Lopez Nov 1 '17 at 19:15
• @RafaelGonzalezLopez $U(t) = \theta(t)-t = \mathcal{O}(t^{1/2} \log t)$ so it is obvious – reuns Nov 1 '17 at 19:16
• @RafaelGonzalezLopez Sure $\int_x^\infty \mathcal{O}(t^{-1/2} \log^2 t) \frac{dt}{t^2} = \mathcal{O}(\int_x^\infty t^{-1/2} \log^2 t \frac{dt}{t^2}) = \mathcal{O}( x^{-1/2} \log^2 x)$. And I meant $\log^2$ everywhere – reuns Nov 1 '17 at 19:28