Integral involving the fractional part function Let $\{\}$ denote the fractional part function and $s>1$ be a real number,  then does the following integral admit a closed-form ?   
$$\int_{0}^{1}\bigg\{\frac{1}{x^s}\bigg\}dx\,$$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\left.\int_{0}^{1}\braces{1 \over x^{s}}\dd x
\,\right\vert_{\ s\ >\ 1} &
\,\,\,\stackrel{x\ =\ t^{\large -1/s}}{=}\,\,\,
{1 \over s}\int_{1}^{\infty}{\braces{t} \over t^{1/s + 1}}\,\dd t
\\[5mm] & =
\sum_{n = 1}^{1}{1 \over n^{1/s}} + {1^{1 - 1/s} \over 1/s - 1} -
\zeta\pars{1 \over s}\label{1}\tag{1}
\\[5mm] & =
1 - {s \over s - 1} - \zeta\pars{1 \over s}
\\[5mm] & = \bbx{-\,{1 \over s - 1} - \zeta\pars{1 \over s}}
\\ &
\end{align}
In \eqref{1}, I used a
Riemann Zeta Identity.
A: By applying the Euler-Maclaurin formula one may get the following relationship :
$$\int_{0}^{1}\bigg\{\frac{1}{x^s}\bigg\}dx=\frac{s}{s-1}+\frac{1}{2s}+\sum_{k\geq1}\frac{B_{2k}}{(2k)!}\bigg(\frac{1}{s}\bigg)^\overline{2k+1}$$
Where $B_k$ denotes the k-th Bernoulli number and $x^\overline{n}$ represents the Pochhammer rising factorial. The question above could be transformed to finding a closed-form of the upper series.   
