# 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 ?

$$\int_{0}^{1}\bigg\{\frac{1}{x^s}\bigg\}dx\,$$

• What do you get from $x = t^{-1/s}$ ? Do you know that $\zeta(s) = \sum_{n=1}^\infty s\int_n^\infty t^{-s-1}dt = s \int_1^\infty \lfloor t \rfloor t^{-s-1}dt$ ? Nov 11, 2018 at 22:54
• @reuns, this substitution gives a similar locked integral. Nov 11, 2018 at 23:08
• I have got already a series representation of the integral in terms of Bernoulli numbers yet the question is if the integral could have a closed-form. Nov 11, 2018 at 23:13
• Can you show what you obtain from $x = t^{-1/s}$ ? .. Nov 11, 2018 at 23:14
• I have left an answer below Nov 11, 2018 at 23:41

$$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ \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.
$$\int_{0}^{1}\bigg\{\frac{1}{x^s}\bigg\}dx=\frac{s}{s-1}+\frac{1}{2s}+\sum_{k\geq1}\frac{B_{2k}}{(2k)!}\bigg(\frac{1}{s}\bigg)^\overline{2k+1}$$
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.