Let $\{\}$ denote the fractional part function and $s>1$ be a real number, then does the following integral admit a closed-form ?
$$\int_{0}^{1}\bigg\{\frac{1}{x^s}\bigg\}dx\,$$
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1$\begingroup$ What do you get from $x = t^{-1/s}$ ? Do you know that $\zeta(s) = \sum_{n=1}^\infty s\int_n^\infty t^{-s-1}dt = s \int_1^\infty \lfloor t \rfloor t^{-s-1}dt$ ? $\endgroup$– reunsNov 11, 2018 at 22:54
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$\begingroup$ @reuns, this substitution gives a similar locked integral. $\endgroup$– Kays TomyNov 11, 2018 at 23:08
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$\begingroup$ 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. $\endgroup$– Kays TomyNov 11, 2018 at 23:13
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$\begingroup$ Can you show what you obtain from $x = t^{-1/s}$ ? .. $\endgroup$– reunsNov 11, 2018 at 23:14
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$\begingroup$ I have left an answer below $\endgroup$– Kays TomyNov 11, 2018 at 23:41
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2 Answers
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.
answered Nov 12, 2018 at 2:02
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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.