Anyone to help me rewrite this : $$\sum _{n=1}^{p}\:\frac{x}{\left(6n-1\right)}\:+\:\frac{x}{\left(6n+1\right)}$$
Where, $p=\left\lfloor \frac{\sqrt{x}-1}{6}\right\rfloor$
to this ( approximations that are easily computed) :
$$xH(x) \approx x (\ln x+\gamma )-\frac{1}{12 x}+\frac{1}{2}$$ Where $\gamma$ is the Euler-Mascheroni constant $\gamma\approx 0.577215665$.
Hint
I think that I should first consider separately $$S_p=\sum _{n=1}^{p}\:\frac{x}{\left(6n-1\right)}\:+\:\frac{x}{\left(6n+1\right)}=\frac{\log (432)-6}{6} x +\frac{H_{p-\frac{1}{6}}+H_{p+\frac{1}{6}}}{6} x $$ and $$p=\left\lfloor \frac{\sqrt{x}-1}{6}\right\rfloor$$ to show a series of discontinuous straight lines changing slope.