# Find a closed form for the constant term

In a previous question, an asymptotic expansion was provided for the weighted divisor summatory function $\displaystyle \frac {d(n)}{n}$:

$$\sum_{n\leq x}\frac{d(n)}{n}=\frac{1}{2}\log^{2}x+2\gamma\log x+\gamma^{2}-2\gamma_{1}+O\left(\frac{1}{\sqrt{x}}\right)$$

Expansions were also provided for the same summatory function when restricted to odd and even divisors. In particular, for the sum restricted to odd divisors it was shown that

$$\sum_{\substack {n\leq x, \\ n \text{ odd}}}\frac{d(n)}{n}=\frac{1}{8} \log^2 x + \left(\frac{\gamma}{2} + \frac{1}{2}\log 2\right) \log x + \frac{\gamma^2}{4} + \gamma \log 2 - \frac{1}{2} \gamma_1 + \mathcal{O}\left(\frac{1}{\sqrt{x}}\right)$$

Because the number of divisors $< \sqrt{n}$ equals that of divisors $\geq \sqrt{n}$ (except for squares), if we add the restriction that, for each $n$, only divisors $\geq \sqrt {n}$ can be counted, the asymptotic result is half of that given by the formula above (at least as regards the main terms).

I would be interested in obtaining a similar asymptotic expansion for the odd summatory function under the restriction that, for any $n$, only divisors $\geq \sqrt{n}$ and $\leq 2\sqrt {n}$ are counted. After some calculations, I got that the asymptotic sum under these conditions becomes considerably simpler, as it is given by

$$\frac {1}{4} \log 2 \log x + K+ \mathcal{O}\left(\frac{1}{\sqrt{x}}\right)$$

where $K$ is a constant term equal to $\approx 0.7059...$. However, I was not able to find any closed form for this constant term. Is there any method to determine it explicitly? Any hint would be very appreciated.

More generally, I found that adding the restriction that only divisors $\geq \sqrt{n}$ and $\leq J\sqrt {n}$ are counted (with $J \geq 1$), the asymptotic sum is $$\frac {1}{4} \log J \log x + K_J+ \mathcal{O}\left(\frac{1}{\sqrt{x}}\right)$$

where $K_J$ is a constant term that varies with $J$. I would also be interested in finding a generalization for the closed form of $K_J$.

• You could search RIES with more digits (if you have) $$\frac{2}{\log (17)}=0.705912$$ $$\sin ^{-1}\left(\sqrt{e}-1\right)=0.705903$$ – Claude Leibovici Aug 25 '14 at 9:58