\xi$) is $$\begin{aligned}
&\frac{1}{n^{2 \alpha}} \int_0^{1 - (1/n)} \frac{n \, d\xi}{n(1 - \xi) \left( \frac{1}{n} + \xi \right)^{2\alpha}} \leq \\
\leq &  \frac{1}{n^{2 \alpha}} \left[ 2 \int_0^{1/2} \frac{d\xi}{\xi^{2 \alpha}} + 2^{2\alpha} \int_{1/2}^{1 - (1/n)} \frac{d\xi}{1 - \xi} \right] =\\
= & \left( \frac{2}{n} \right)^{2\alpha} \left[ \frac{1}{1-2\alpha} + \log \frac{n}{2} \right].
\end{aligned} \label{eq:bounda2.1}$$ The second integral in is equal to $$\begin{aligned}
&\frac{1}{n^{2\alpha}} \int_{1 + (2/n)}^{\infty} \frac{d\eta}{(\eta - 1)\left( \frac{1}{n} + \eta \right)^{2\alpha}} \leq \\
\leq &  \frac{1}{n^{2 \alpha}} \left[  \int_{1 + (2/n)}^{2} \frac{d\eta}{\eta - 1} +  \int_{2}^{\infty} \frac{d\eta}{(\eta - 1)^{2\alpha + 1}} \right] =\\
= & \frac{1}{n^{2\alpha}} \left[ \log \frac{n}{2} + \frac{1}{2\alpha} \right].
\end{aligned} \label{eq:bounda2.2}$$ If we collect the inequalities , we get (with $2^{2\alpha} \leq 2$) $$\begin{aligned}
&\begin{aligned}
\mu &\leq \frac{2}{3} \frac{1}{n^{2\alpha}} \left[ 9 + \frac{3}{2} + \frac{2}{1 - 2\alpha} + 3 \log \frac{e n}{2} + \frac{1}{2\alpha}  \right] \leq \\
&\leq \frac{1}{n^{2\alpha}} [M(\alpha) + 2 \log n].
\end{aligned} \label{eq:muest}\\
&\text{ where } M(\alpha) = 6 + \frac{4/3}{1- 2 \alpha} + \frac{1}{3\alpha}. \label{eq:malphdef}\end{aligned}$$ With and we come to .

Of course, the constant factors in inequalities – are not sharp but we get some idea on their magnitude. If $\alpha = {\displaystyle}\frac{1}{4}$ we have $$\begin{gathered}
M\left( \frac{1}{4} \right) = 6 + 