So I know that If $A(x) = \sum_{n \leq x} a(n)$, and $\phi(x)$ has a continuous derivative on $[X_1, X_2]$, where $0 \leq X_1 < X_2 \in \mathbb{R}$, then
$$\sum_{X_1 < n \leq X_2} a(n)\phi(n) = A(X_2)\phi(X_2) – A(X_1)\phi(X_1)- \int_{X_1}^{X_2} A(x)\phi'(x)\,\mathrm dx\tag1\label{eq1}$$
I also know that
$$\sum_{n \leq x} d(n) = x(\log x + 2e – 1) + \mathcal O(\sqrt{x})\tag2\label{eq2}$$
So from $\eqref{eq1}, \eqref{eq2}$ letting $a(n) = d(n)$ and $\phi(n) = 1/n$, I get
$$\begin{align}\sum_{n \leq x} d(n)/n &= \frac{1}{x} \sum_{n \leq x}d(n) + \int_{1}^x d(t)/t^2 \, \mathrm dt
\\&= (\log x + 2e – 1) + \mathcal O(\sqrt{x})/x + \int_{1}^x d(t)/t^2 \,\mathrm dt
\\&= (\log x + 2e – 1) + \mathcal O(1/\sqrt{x}) + \int_{1}^x d(t)/t^2 \,\mathrm dt\end{align}$$
but I am not sure where to go from here. I was thinking of rewriting $\int_{1}^x d(t)/t^2 \,\mathrm dt$ as
$$\begin{align}\int_{1}^x \frac{d(t)}{t^2} \, \mathrm dt &=\int_1^x \frac{\log(t)}{t}\,\mathrm dt + \int_{1}^x \frac{d(t) – t\log(t)}{t^2} \, \mathrm dt
\\&= \frac{1}{2}\log^2x + \int_{1}^x \frac{d(t) – t\log(t)}{t^2} \, \mathrm dt
\\&= \frac{1}{2}\log^2x + \int_{1}^\infty \frac{d(t) – t\log(t)}{t^2} \, \mathrm dt – \int_{x}^\infty \frac{d(t) – t\log(t)}{t^2} \, \mathrm dt\end{align}$$
Would either of these two integrals converge to a constant?
Best Answer
\begin{align} \int_{1}^x \frac{d(t)}{t^2} \, \mathrm dt &=\int_1^x \frac{(\log(t) +2e -1)t+ \mathcal{O}(\sqrt{t})}{t^2}\ dt \\&= \frac{1}{2}\log^2x + (2e-1) \log x + \int_{1}^x \mathcal{O}(t^{-\frac{3}{2}}) dt, \end{align} and $\int_{1}^\infty\mathcal{O}(t^{-\frac{3}{2}})dt$ is finite.