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If $d(n) = \sum_{d|n}1$ is the divisor function, it is a fact that $d$ has summatory behaviour \begin{equation} \sum_{n \leq x} d(n) = x \log x + (2 \gamma - 1)x + O(\sqrt{x}). \end{equation}

I'm interested in the behaviour of $\sum_{n \leq x} \frac{d(n)}{n}$ for large $x$. Now, this answer uses the Dirichlet Hyperbola Method, which I understand, to get the result \begin{equation} \sum_{n \leq x} \frac{d(n)}{n} = \frac{1}{2} \log^2 x + 2 \gamma \log x + \gamma^2 -2\gamma_1 + O\left(\frac{1}{\sqrt x}\right). \end{equation} However, if I try using Abel summation (motivated by this similar question), I get the bound \begin{equation} \sum_{n \leq x} \frac{d(n)}{n} = \frac{1}{x}\left(x \log x + (2 \gamma - 1) x + O(\sqrt x)\right) -1 + \int_1^x\frac{t \log t + (2 \gamma - 1) t + O(\sqrt t)}{t^2}dt \end{equation} \begin{equation} =\frac{1}{2}\log^2 x + 2 \gamma\log x + 2(\gamma - 1) + O\left(\frac{1}{\sqrt x} \right). \end{equation}

Now the numerical constants $\gamma^2-2\gamma_1$ and $2(\gamma-1)$ are not equal; more worryingly I see no way that my approach could have yielded $\gamma_1$ leading me to believe the idea was fundamentally flawed. Would appreciate if someone could aide my understanding.

Chris
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  • Perhaps $\gamma_1=-\gamma ^2-\int_0^{\infty} \left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right) e^{-x} \log (x) , dx$ provides a clue. – Steven Clark May 26 '22 at 21:55
  • An interesting relation. However per the selected answer it seems the best error I can do is $O(1)$ so unfortunately I wouldn't be able to use this. – Chris May 27 '22 at 09:28

1 Answers1

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You probably made the following mistake:

$$\int_1^x \frac{O(\sqrt{t})}{t^2}\,dt = O\left(\frac{1}{\sqrt{x}}\right)$$

That's not right.

Observe that $$\int_1^x \frac{O(\sqrt{t})}{t^2}\,dt = O\left(\int_1^x \frac{\sqrt{t}}{t^2}\,dt\right)=O\left(2-\frac{2}{\sqrt{x}}\right) = O(1)$$

which means that using Abel summation, the best approximation you can get is $$\sum_{n\le x} \frac{d(n)}{n} = \frac{1}{2}\log^2 x + 2 \gamma\log x + O(1)$$

jjagmath
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