$\int_{0<x<y<z<w<1} \left\{\frac{x}{y}\right\}\left\{\frac{y}{z}\right\}\left\{\frac{z}{w}\right\}\left\{\frac{w}{x}\right\}$

Question

I was trying to integrate this : $\displaystyle \int\limits_{0\lt x\lt y\lt z\lt w\lt 1} \left\{\dfrac{x}{y}\right\}\left\{\dfrac{y}{z}\right\}\left\{\dfrac{z}{w}\right\}\left\{\dfrac{w}{x}\right\}$

where $\{.\}$ denotes the Fractional Part function. What I did was :

$\displaystyle \begin{align}I &= \displaystyle \int\limits_{0\lt x\lt y\lt z\lt w\lt 1} \dfrac{x}{y}\dfrac{y}{z}\dfrac{z}{w}\left\{\dfrac{w}{x}\right\} \\ &= \int_{w=0}^1\int_{x=0}^{w}\int_{y=x}^{w}\int_{z=x}^{w} \dfrac{x}{y}\dfrac{y}{z}\dfrac{z}{w}\left\{\dfrac{w}{x}\right\} \; dx\; dy\; dz\; dw \\ &=\int_{w=0}^1 \int_{x=0}^{w} \dfrac{(w-x)^2 x}{w}\left\{\dfrac{w}{x}\right\}\; dx\; dw \\ &=\int_0^1 \int_0^1 w^2 (1-t)^2 t\left\{\dfrac{1}{t}\right\}\; dt\; dw\quad \text{By substitution }x=wt \\ &=\dfrac{1}{3}\int_0^1 t(1-t)^2 \left\{\dfrac{1}{t}\right\}\; dt\end{align}$

Now I could easily evaluate the last integral and if there isn't a silly mistake it comes out to be $I=\dfrac{2\zeta(3)}{9}-\dfrac{\zeta(4)}{12}-\dfrac{\zeta(2)}{6}+ \dfrac{1}{9}$

I am here cuious that did I manipulated the limits well ? i.e breaking $0<x<y<z<w<1$ into four different integrations with proper limits. Were the limits all right ?

Answer 1

This is a comment addressing OP's real question how to decide the integration limit.
It is simply too long to fit into the comment box.

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The bounds of $z$ can be $\int_{z=0}^w$, $\int_{z=x}^w$ or something else. Which one to use depends on the order of integration. As an example, consider following strange order of integration $ dw dy dx dz$.

1. The outermost layer of integration is $z$. Since this is the outermost layer, the other variables $x,y,w$ have been handled by the inner layer of integration. The only constraint of it is $0 < z < 1$. This means your integral begin with a $\int_{z=0}^1$.

2. Next you fix $z$, the condition $0 < x < y < z < w$ split into two sub-conditions $0 < x < y < z$ and $z < w < 1$. Since you want to integrate $x$, only the first sub-condition matter. Since $y$ has been handled by inner layer of integral, the constraint on $x$ with be $0 < x < z$. This means the first two layer of integral should be $\int_{z=0}^1\int_{x=0}^z$.

3. Following that, $z$ and $x$ are fixed and you want to integrate over $y$. The relevant constraint on $y$ is $x < y < z$. The first three layer of integral should be $\int_{z=0}^1 \int_{x=0}^z\int_{y=x}^z$.

4. Finally, $z$, $x$, $y$ are fixed and we are going to integrate over $w$. The only constraint remain is $z < w < 1$, The full integral should be $\int_{z=0}^1\int_{x=0}^z\int_{y=x}^z\int_{w=z}^1 (\cdots)dw dy dx dz$

In short, you can determine the correct integral limits by enforcing the constraint layer from layer. Start from the outermost layer and working inwards.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, I'll use Iverson Brackets $\ds{\pars{~\mbox{namely,}\ \bracks{\cdots}~}}$ which are quite efficient whenever a cumbersome restriction is present.

\begin{align} I & \,\,\,\substack{\mbox{def.} \\[1mm] \ds{\equiv}} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \braces{x \over y}\braces{y \over z}\braces{z \over w}\braces{w \over x} \bracks{0 < x < y < z < w < 1}\dd x\,\dd y\,\dd z\,\dd w \\[5mm] & = \int_{-\infty}^{1}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \int_{0}^{\infty}{x \over y}\,{y \over z}\,{z \over w}\,\braces{w \over x} \bracks{x < y < z < w}\dd x\,\dd y\,\dd z\,\dd w \\[5mm] & = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{0}^{\infty}\int_{-\infty}^{1} {1 \over w/x}\,\braces{w \over x} \bracks{x < y < z < x\,{w \over x}}x\,{\dd w \over x}\,\dd x\,\dd y\,\dd z \\[5mm] & = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{0}^{\infty}x\int_{-\infty}^{1/x} {\braces{w} \over w}\bracks{x < y < z < xw}\dd w\,\dd x\,\dd y\,\dd z \\[5mm] & = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{0}^{\infty}x\int_{-\infty}^{\infty} \bracks{w < {1 \over x}}{\braces{w} \over w}\bracks{x < y < z}\bracks{z < xw} \dd w\,\dd x\,\dd y\,\dd z \\[5mm] & = \int_{0}^{\infty}x\int_{-\infty}^{\infty}\bracks{w < {1 \over x}} {\braces{w} \over w}\int_{-\infty}^{\infty}\bracks{z < xw} \int_{-\infty}^{\infty}\bracks{x < y < z}\dd y\,\dd z\,\dd w\,\dd x \\[5mm] & = \int_{0}^{\infty}x\int_{-\infty}^{\infty}\bracks{w < {1 \over x}} {\braces{w} \over w}\int_{-\infty}^{\infty}\bracks{z < xw} \bracks{x < z}\int_{x}^{z}\dd y\,\dd z\,\dd w\,\dd x \\[5mm] & = \int_{0}^{\infty}x\int_{-\infty}^{\infty}\bracks{w < {1 \over x}} {\braces{w} \over w}\int_{-\infty}^{\infty}\bracks{x < z < xw} \pars{z - x}\dd z\,\dd w\,\dd x \\[5mm] & = \int_{0}^{\infty}x\int_{-\infty}^{\infty}\bracks{w < {1 \over x}} {\braces{w} \over w}\bracks{x < xw}\int_{x}^{xw} \pars{z - x}\dd z\,\dd w\,\dd x \\[5mm] & = \int_{0}^{\infty}x\int_{-\infty}^{\infty}\bracks{1 < w < {1 \over x}} {\braces{w} \over w} \pars{{1 \over 2}\,w^{2}x^{2} - {1 \over 2}\,x^{2} - wx^{2} + x^{2}}\dd w\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{\infty}x^{3}\int_{-\infty}^{\infty} \bracks{1 < w < {1 \over x}}{\braces{w} \over w}\pars{w - 1}^{2} \,\dd w\,\dd x \\[5mm] & = {1 \over 2}\int_{-\infty}^{\infty} {\braces{w} \over w}\pars{w - 1}^{2}\int_{0}^{\infty}x^{3} \bracks{x < wx < 1}\,\dd x\,\dd w \\[5mm] & = {1 \over 2}\int_{-\infty}^{\infty} {\braces{w} \over w}\pars{w - 1}^{2}\bracks{w > 1}\int_{0}^{1/w}x^{3} \,\dd x\,\dd w = {1 \over 8}\int_{1}^{\infty} {\pars{w - 1}^{2} \over w^{5}}\,\pars{w - \left\lfloor w\right\rfloor}\dd w \\[5mm] & = { 1 \over 24} - {1 \over 8}\sum_{n = 1}^{\infty}\int_{n}^{n + 1}{\pars{w - 1}^{2} \over w^{5}}\,n\,\dd w = \bbx{{1 \over 24} - {1 \over 16}\,\zeta\pars{2} + {1 \over 12}\,\zeta\pars{3} - {1 \over 32}\,\zeta\pars{4}} \\[5mm] & = \bbx{{1 \over 24} - {\pi^{2} \over 96} + {1 \over 12}\,\zeta\pars{3} - {\pi^{4} \over 2880}} \approx 0.00520709503181230\ldots \\ & \end{align}

I already performed a simulation, up to $\ds{16,777,215}$ quadruplets $\ds{\pars{x,y,z,w}}$, which agrees with the above analytic result. Such simulation 'generates' the value $\ds{0.00520\color{#f00}{943}}$.