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I need help in evaluating the following integrals involving the fractional part: \begin{equation} I_{k} = \int_{t = 0}^{1}\int_{y = y_{k}}^{y_{k+1}}\int_{x = x_{k}}^{x_{k+1}} \left\{\dfrac{y}{t}\right\}\left\{\dfrac{t}{x}\right\} dt dy dx \end{equation} where $x_{k} = y_{k} = \frac{k}{n}$, $k = 0,\dots,n-1 $ and $n \geqslant 2$. Here, $\{z\} $ denotes the fractional part of $z$ $(z > 0)$, i.e. $\{z\} = z - \lfloor z\rfloor$ where $\lfloor z \rfloor$ is the greatest integer less than or equal to $z$. If we note $h = \frac{1}{n}$, then the integrals could be expressed in terms of $h$ (and $k$, obviously).

Thank you.

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$$\begin{eqnarray*} I_k &=& \int_{0}^{1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\left\{\frac{y}{t}\right\}\left\{\frac{t}{x}\right\}\,dt\,dy\,dx \\ &=& \frac{1}{n^2}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left\{\frac{k+y}{k+t}\right\}\left\{\frac{k+t}{nx}\right\}\,dt\,dy\,dx\end{eqnarray*} $$

now the argument of the first fractional part is always between $\frac{k}{k+1}$ and $\frac{k+1}{k}$, so: $$\begin{eqnarray*} I_k &=& \frac{1}{n^2}\int_{0}^{1}\iint_{(0,1)^2}\frac{|y-t|}{k+t}\left\{\frac{k+t}{nx}\right\}\,dt\,dy\,dx\\&=&\frac{1}{n^3}\int_{0}^{n}\int_{0}^{1}\frac{t^2-t+\frac{1}{2}}{k+t}\left\{\frac{k+t}{x}\right\}\,dt\,dx\end{eqnarray*} $$ by exploiting $\int_{0}^{1}|y-t|\,dy = t^2-t+\frac{1}{2}$. Now you may split the integration range according to $0\leq k+t\leq nx$, $nx\leq k+t\leq 2nx$ and so on. I think you can take it from here.

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  • $\begingroup$ Hi Jack, sorry for my so late response. So, I am still in need to find a solution. I tried yours, continuing from you left, but everything seems to go around a divergent integral term. To explain: if j$ \leq \left\{\frac{k+t}{x}\right\} < $ j+1, then the expression under integrals splits in two quantities, first one having a factor 1/x another factor the polynom in t, i.e. $\frac{t^2}-t+1/2}{x}$; so, we will obtain a divergent integral $\int_{0}^{1} (1/x) dx$, as a factor in the first quantity. \\ May be you could help in solving the integral. Thanks in advance. $\endgroup$ – Dea Jun 13 '16 at 2:57
  • $\begingroup$ correction to previous comment:Hi Jack, sorry for my so late response. So, I am still in need to find a solution. I tried yours, continuing from you left, but everything seems to go around a divergent integral term. To explain: if j$ \leq \left\{\frac{k+t}{x}\right\} < $ j+1, then the expression under integrals splits in two quantities, first one having a factor 1/x another factor the polynom in t, i.e. $\frac{t^2-t+1/2}{x}$; so, we will obtain a divergent integral $\int_{0}^{1} (1/x) dx$, as a factor in the first quantity. \\ May be you could help in solving the integral. Thanks in advance. $\endgroup$ – Dea Jun 13 '16 at 3:06

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