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It is well known (see here, for example) that we have

$$ \psi\left(\frac{1}{2},T\right)=\sum_{p\leq T}\frac{1}{p}=\log(\log T)+A+O\left(\frac{1}{\log T}\right), $$ where $\psi(\sigma,T)=\sum_{p\leq T}p^{-2\sigma}$, $p$ denotes a prime number, and $A$ is some constant.

I'm interested in which $\sigma>\frac{1}{2}$ the sum has essentially the same order (note that I'm not asking about convergence/divergence; clearly it converges, but $\log(\log T)$ might still well approximate the sum).

What I've done

If we put $\sigma=\frac{1}{2}+\omega(T)$, we have $$ \psi(\sigma,T)=\sum_{p\leq T}\frac{p^{-2\omega(T)}}{p}=\sum_{p\leq T}\frac{1-2\omega(T)\log(p)+\cdots}{p}. $$ Thus, if $\omega(T)$ is "nice" enough, we have $$ \psi(\sigma,T)=\log(\log T)+2\omega(T)\log(T)+O(\omega(T)).\tag{$\dagger$} $$ Thus, it would seem that as long as $\omega(T)=O\left(\frac{\log(\log T)}{\log T}\right)$, the order of magnitude is essentially the same.

My Question

Is this correct? Can it be streamlined/improved? I'm staring at a paper of Atle Selberg that makes it seem like $\omega(T)=O((\log T)^{-\delta})$ for some "fixed positive" $\delta$ should hold. While I think at the moment that my work is correct, I'm not convinced it is what he had in mind when he wrote the passage.

($\dagger$) I realize I am sweeping some details under the rug here; if necessary, I can post them, but it seems fruitless to do so at the moment.

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Clayton
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  • Be careful sweeping details under the rug as something is not right with how you truncated the exponential series and put in a big-O term. In particular, we should not have $O(\omega(T))$ on the right hand side. For monotonic $\omega$, if $$\omega(t)=o\left(\frac{1}{\log t}\right)$$ then you will have a main term of $\log \log x$. If $$\omega(t)=\frac{c}{\log t}\left(1+o(1)\right),$$ then you will have a main term of $e^{-2c} \log \log x$, and for $\omega(t)$ larger than this, the sum will not behave like $\log \log x$ anymore. – Eric Naslund Jul 23 '15 at 14:53
  • To see this, note that the relevant integral is $$\int_{2}^{x}\frac{t^{-2\omega(x)}}{t\log t}dt,$$ and it satisfies $$\int_{2}^{x}\frac{t^{-2\omega(t)}}{t\log t}dt\geq\int_{2}^{x}\frac{t^{-2\omega(x)}}{t\log t}dt\geq x^{-2\omega(x)}\log\log x.$$ – Eric Naslund Jul 23 '15 at 14:58
  • @EricNaslund: You're right that my estimate wasn't correct. I fleshed out the details for how I was thinking about it and found that my hoped-for estimate didn't work out. I'll keep working on it, but thank you. – Clayton Jul 23 '15 at 17:28

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