Asymptotic for $\sum_{n \le x}\frac{d(n)\log n}{n}$

Let $H(x)=\displaystyle\sum_{n \le x}\cfrac{d(n)\log n}{n}$. I want to find asymptotic for $H(x)$, come up to some point and need help to finish it.

First, recall the Abel's summation formula:

Given an arithmetical function $a(x)$ define $A(x):= \displaystyle\sum_{n \le x}a(x)$ and assume that $f$ is a function which has continuous derivative on $[1,x]$.

Then, $\displaystyle\sum_{n \le x}a(n)f(n) = A(x)f(x)-\int_{1}^{x}A(t)f'(t) dt$.

Now we can choose $d(n)$ as $a(n)$ and $f(x)$ as $\cfrac{\log x}{x}$. We can say that $A(x)= \displaystyle \sum_{n \le x}d(n)$.

We know that $A(x) = x \log x +(2C-1)x +O(\sqrt{ x})$ or $A(x)=x \log x + O(x),$ rougly. And say $C^*=2C-1$ in short.

Then, $H(x)=\displaystyle\sum_{n \le x}\cfrac{d(n)\log n}{n} = A(x)\log(x) + \int_{1}^{x}A(t)\cfrac{1- \log t}{t^2}dt$.

$H(x)=x\log^2(x) +C^* \log x +O(x^{3/2}) + \displaystyle\int_{1}^{x}(t\log t +C^*+O(\sqrt t))\cfrac{1-\log t}{t^2}dt$

$=x\log^2(x) +C^* \log x +O(x^{3/2}) +\cfrac{\log^3 x}{3} - \cfrac{\log^2x}{2} + \displaystyle\int_{1}^{x}O(\sqrt t))\cfrac{1-\log t}{t^2}dt$

I have an idea for finding $\displaystyle\int_{1}^{x}O(\sqrt t))\cfrac{1-\log t}{t^2}dt$ but want to ask, what can I do now? Is there any mistakes? Thank you.

• There are many mistakes in the asymptotics for $H(x)$, which is $\sim P(\log x)$ (with $P$ a polynomial of degree $3$ ?) not $x P(\log x)$. – reuns Nov 15 '17 at 4:18
• Why only one $\log x$ multiplier is enough? and cannot we multiply as I did? – Ninja Nov 15 '17 at 6:53
• allo ${}{}{}{}$ – reuns Nov 17 '17 at 2:59

Using the hyperbola method $$D(k) = \sum_{n=1}^k d(n) = k(C+\log k+\mathcal{O}(k^{-1/2}))$$ Summing by parts $$H(k)=\sum_{n =1}^k d(n) \frac{\log n}{n} = \frac{\log k}{k}D(k)+ \sum_{n =1}^{k-1} D(n) (\frac{\log n}{n}-\frac{\log (n+1)}{(n+1)})$$ Where $$\frac{\log n}{n}-\frac{\log (n+1)}{(n+1)} =\int_n^{n+1} \frac{\log x-1}{x^2} dx = \frac{\log n-1}{n^2}+\mathcal{O}(\frac{\log n}{n^{3}})$$ Thus $$H(k)= \frac{\log k}{k} k(C+\log k+\mathcal{O}(k^{-1/2}))\! +\!\!\sum_{n =1}^{k-1}n(C+\log n+\mathcal{O}(n^{-1/2})) (\frac{\log n-1}{n^2}+\mathcal{O}(\frac{\log n}{n^{3}}))$$ $$= P(\log k)+B+O(k^{-1/2}\log k)$$ with $P$ a polynomial of degree $3$ (due to the term $\sum_{n=1}^k \frac{\log^2 n }{n} \sim \log^3 k$) and $B$ is the limit of the remainder convergent sequences.
Note $\sum_{n=1}^\infty d(n) n^{-s-1}\log n = - 2\zeta'(s+1)\zeta(s+1)$ thus $P(\log x) +B= \text{Res}(- 2\zeta'(s+1)\zeta(s+1)\frac{x^s}{s},0)$
• @Ninja ?? What do you not understand in my answer ? Abel summation formula is the same as summing by parts then replacing $\frac{\log n}{n}-\frac{\log (n+1)}{(n+1)}$ by $\int_n^{n+1} \frac{\log x-1}{x^2} dx$. What I wrote is exactly what you should have done. – reuns Nov 15 '17 at 21:49