Integral of FractionalPart function

Question

My learning curve began with a problem on Brilliant:

$\large \int_0^1\int_0^1\lbrace{\frac{x^3}{y}\rbrace}dxdy=\frac{A}{B}-\frac{\gamma}{C}$

I recognized the problem with the lower bound on y being zero. Since for both x and y we are going from 0 to 1, the confusion between the Mathematica/GeoGebra definition for {x} , ie frac(x) and the Graham et al. definition, was not an issue on this question. Then I went ahead and worked out the inside integral to be $\frac{1}{4y}$. With trepidation, and not sure even that step was correct, I was looking at:

$\int_0^1\frac{1}{4y}dy$

That's where I started to spin my wheels, get frustrated and confused. More research followed. More frustration ensued. I then tried to go back and attempt numerical analysis from square one with a double integral tool on GeoGebra.org ..... The best I did there was finding a lower bound of 10 and an upper bound of 20 for the final answer but I am not confident even that's true. I would be happy if all I got was a numerical result accurate to 8 decimal digits but I just can't handle the discontinuity at y=0 I guess. That the Euler-Mascheroni constant .57721566 is involved in the closed form template for the result is very intriguing to me. Currently, the Brilliant site has no posted solutions yet, or I would have put in 3 guesses and got to see it days ago. The tutorial Wikis on Brilliant helped me understand how non-trivial the fractionalPart function is! Sorry about the long story.

Answer 1

Hint:

• $0\le x^3 <y \le 1 \implies \left \{ \dfrac{x^3}{y} \right \}=\dfrac{x^3}{y}$

\begin{align*} \int_{0}^{1} \int_{x^3}^{1} \left \{ \frac{x^3}{y} \right \} \, dy \, dx &= \int_{0}^{1} \int_{x^3}^{1} \frac{x^3}{y} \, dy \, dx \\ &= \frac{3}{16} \end{align*}

• $0\le \dfrac{x^3}{n+1} <y \le \dfrac{x^3}{n} \le \dfrac{1}{n} \implies \left \{ \dfrac{x^3}{y} \right \}=\dfrac{x^3}{y}-n$

\begin{align*} \int_{0}^{1} \int_{\frac{x^3}{n+1}}^{\frac{x^3}{n}} \left \{ \frac{x^3}{y} \right \} \, dy \, dx &= \int_{0}^{1} \int_{\frac{x^3}{n+1}}^{\frac{x^3}{n}} \left( \frac{x^3}{y}-n \right) dy \, dx \\ &= \frac{1}{4} \left( \ln \frac{n+1}{n}-\frac{1}{n+1} \right) \end{align*}

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#ffe,15px]{\ds{% \int_{0}^{1}\int_{0}^{1}\braces{x^{3} \over y}\,\dd x\,\dd y}} = \int_{0}^{1}x^{3}\int_{0}^{1/x^{3}}\braces{1 \over y}\,\dd y\,\dd x = \int_{0}^{1}x^{3}\int_{x^{3}}^{\infty}{\braces{y} \over y^{2}}\,\dd y\,\dd x \\[5mm] = &\ \int_{0}^{\infty}{\braces{y} \over y^{2}}\int_{0}^{1}x^{3} \bracks{x < y^{1/3}}\,\dd x\,\dd y = \int_{0}^{1}{\braces{y} \over y^{2}}\int_{0}^{y^{1/3}}x^{3}\,\dd x\,\dd y + \int_{1}^{\infty}{\braces{y} \over y^{2}}\int_{0}^{1}x^{3}\,\dd x\,\dd y \\[5mm] = &\ {1 \over 4}\int_{0}^{1}{y \over y^{2/3}}\,\dd y + {1 \over 4}\int_{1}^{\infty}\pars{{1 \over y} - {\left\lfloor y\right\rfloor \over y^{2}}}\,\dd y = {3 \over 16} + {1 \over 4}\sum_{n = 1}^{\infty} \int_{n}^{n + 1}\pars{{1 \over y} - {n \over y^{2}}}\,\dd y \\[5mm] = &\ {3 \over 16} + {1 \over 4}\sum_{n = 1}^{\infty} \bracks{\ln\pars{1 + n} - \ln\pars{n} + {n \over n + 1} - 1} \\[5mm] = &\ {3 \over 16} + {1 \over 4}\lim_{N \to \infty}\bracks{\sum_{n = 1}^{N} \ln\pars{1 + n} - \sum_{n = 1}^{N}\ln\pars{n} - \sum_{n = 1}^{N}{1 \over n + 1}} \\[5mm] = &\ {3 \over 16} + {1 \over 4}\lim_{N \to \infty}\bracks{\sum_{n = 2}^{N + 1} \ln\pars{n} - \sum_{n = 1}^{N}\ln\pars{n} - \sum_{n = 2}^{N + 1}{1 \over n}} = {3 \over 16} + {1 \over 4}\lim_{N \to \infty}\bracks{\ln\pars{N + 1} + 1 - \sum_{n = 1}^{N + 1}{1 \over n}} \\[5mm] = &\ \bbx{\ds{{7 \over 16} - {1 \over 4}\,\gamma}} \end{align}