From Theorem 4.9 of Apostol Introduction to Analytic Theory, p 88
$$\tilde{\psi} \left({x}\right) =
\tilde{\psi}_{1} \left({x}\right) = \sum_{n=1}^{x} \frac{\Lambda \left({n}\right)}{n} \sim \log \left({x}\right) + O \left({1}\right)$$
Where $\Lambda \left({n}\right)$ is von Mangoldt's function. I also see that this asymptotic form is a consequence of Shapiro’s Tauberian Theorem. Now from Xavier Gourdon and Pascal Sebah ("Collection of formulae for Euler’s constant 𝛾") (http://numbers.computation.free.fr/Constants/Gamma/gammaFormulas.pdf) Equation 7 we have $$\gamma = – \frac{1}{2} \sum_{n = 1}^{\infty} \frac{\Lambda \left({n}\right) – 1}{n}$$.
If we do partial sums, it looks like I can write
$$\tilde{\psi}_{1} \left({x}\right) = \sum_{n = 1}^{x} \frac{\Lambda \left({n}\right)}{n} \sim
– 2\, \gamma + {H}_{x} \sim
\log \left({x}\right)
– \gamma
+ O \left({\frac{1}{x}}\right).$$
Numerical testing verifies this very accurately. I do not see this relation in my literature searches. I see Apostol's answer everywhere. What is the error in my estimation if any?
The numerical table for $n \in \left\{{1, 10, 10^2, 10^3, 10^4, 10^5}\right\}$ below
$\tilde{\psi} \left({x}\right)$" />
Best Answer
You are using wrong the symbol for asymptotic equivalence. Stating, for example, that $H_x\sim \log x + \gamma + O(1/x)$ makes no sense since asymptotic equivalence doesn't care for the terms of smaller order. For example $H_x \sim \log x \sim \log x+\gamma \sim \log x + 17 \sim \log x + \log\log x$.
From the equation $$\gamma = -\frac{1}{2} \sum_{n=1}^\infty \frac{\Lambda(n)-1}{n}$$ you may conclude
$$\sum_{n=1}^x \frac{\Lambda(n) - 1}{n} = -2\gamma+o(1)$$
so $$\sum_{n=1}^x \frac{\Lambda(n)}{n} = H_x - 2\gamma+o(1) = \log x + \gamma+O(1/x)-2\gamma+o(1) = \log x -\gamma + o(1)$$
(Notice that the best you can do for the error term is $o(1)$ which is a lot weaker than $O(1/x)$).