Today I'm trying to defeat another fractional part integral but it seems quite difficult… That is $$\int_0^1 \int_0^1 \left\{\frac{x+y}{x-y}\right\}\,\mathrm{d}x\,\mathrm{d}y$$ Probably, a way to solve is noticing that for the definition of fractional part function $$\left\{\frac{x+y}{x-y}\right\}=\left\{1+\frac{2y}{x-y}\right\}=\left\{\frac{2y}{x-y}\right\}$$ and it seems somewhat easier. I suppose that the answer is in evaluating the fractional integrand but I have some difficulties with it and the substitutions that I tried weren't very useful. Any suggestion?
Another double fractional part integral
definite integralsfractional-partintegrationmultivariable-calculus
Related Solutions
It is helpful to derive the asymptotic expansion of $g$ first. We can use the binomial series to find \begin{align} g(n) &= \sum \limits_{k=2}^n k \sqrt{1-k^{-2}} = \sum \limits_{k=2}^n k \sum \limits_{j=0}^\infty {1/2\choose j} (-k^{-2})^j \\ &= \frac{n(n+1)}{2} - 1 - \frac{H_n}{2} + \frac{1}{2} + \sum \limits_{j=2}^\infty {1/2\choose j} (-1)^j \sum \limits_{k=2}^n k^{1-2j} \end{align} with the harmonic numbers $H_n$. The monotone convergence theorem now yields the asymptotic equivalence $$ g(n) \sim \frac{n(n+1)}{2} - \frac{H_n}{2} + c_g + \mathcal{o}(1)$$ as $n \to \infty$ . The constant term can be written as $$ c_g = - \frac{1}{2} + \sum \limits_{j=2}^\infty {1/2\choose j} (-1)^j [\zeta(2j-1) - 1] = \sum \limits_{k=2}^\infty \left(\sqrt{k^2-1} - k + \frac{1}{2k}\right) \, ,$$ which agrees with the integral representation after using the series expansion of $I_1$.
In order to find $i$ we use the substitution $x = t - \sqrt{t^2-1}$ : \begin{align} i &= \int \limits_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\} \, \mathrm{d} x = \int \limits_1^\infty \{t\} \left(\frac{t}{\sqrt{t^2-1}}-1\right) \, \mathrm{d} t \\ &= \sum \limits_{n=1}^\infty \int \limits_n^{n+1} (t-n) \left(\frac{t}{\sqrt{t^2-1}}-1\right) \, \mathrm{d} t \\ &= \frac{1}{2} \sum \limits_{n=1}^\infty \left[\ln\left(\sqrt{(n+1)^2-1}+n+1\right) - \ln\left(\sqrt{n^2-1}+n\right)\right. \\ &\phantom{= \frac{1}{2} \sum \limits_{n=1}^\infty\left[\right.} \left.- (n+1)\sqrt{(n+1)^2-1} + n \sqrt{n^2-1} + 2\sqrt{(n+1)^2 - 1} - 1 \right] \, . \end{align} The remaining series is (mostly) telescoping and we obtain \begin{align} i &= \frac{1}{2} \lim_{N \to \infty} \left[\ln\left(\sqrt{N^2-1} + N\right) - N \sqrt{N^2-1} + 2 g(N) - N + 1\right] \\ &= \frac{1}{2} \lim_{N \to \infty} \left[\ln\left(1+\sqrt{1-N^{-2}}\right) + \ln(N) - H_N + N \left(N+1 - \sqrt{N^2-1} - 1\right) + 2 c_g + 1\right] \\ &= \frac{1}{2} \left[\ln(2) - \gamma + \frac{1}{2} + 2 c_g + 1\right] \\ &= \frac{3}{4} + \frac{\ln(2)-\gamma}{2} + c_g \, . \end{align}
Substitute $x=1/t$ and then \begin{align*} &\int_1^{ + \infty } {\sqrt {\frac{{\left\{ t \right\}}}{{1 - \left\{ t \right\}}}} \frac{1}{{t(t - 1)}}dt} = \sum\limits_{n = 1}^\infty {\int_n^{n + 1} {\sqrt {\frac{{\left\{ t \right\}}}{{1 - \left\{ t \right\}}}} \frac{1}{{t(t - 1)}}dt} } \\ & = \sum\limits_{n = 1}^\infty {\int_0^1 {\sqrt {\frac{{\left\{ {n + s} \right\}}}{{1 - \left\{ {n + s} \right\}}}} \frac{1}{{(n + s)(n + s - 1)}}ds} } \\ &= \int_0^1 {\sqrt {\frac{s}{{1 - s}}} \left( {\sum\limits_{n = 1}^\infty {\frac{1}{{(n + s)(n + s - 1)}}} } \right)ds} \\ &= \int_0^1 {\sqrt {\frac{s}{{1 - s}}} \left( {\sum\limits_{n = 1}^\infty {\left( \frac{1}{{n + s - 1}}-\frac{1}{{n + s}}\right)} } \right)ds} \\ & = \int_0^1 {\frac{1}{{\sqrt {1 - s} }}\frac{1}{{\sqrt s }}ds} = B\left( {\tfrac{1}{2},\tfrac{1}{2}} \right) = \frac{{\Gamma \left( {\frac{1}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma (1)}} = \pi . \end{align*} So we used the integral representation of the beta function.
Best Answer
The integrand has interesting behavior around $y=x$, so let's split the integral there into two pieces. Let's convert the bottom piece into an integral in polar coordinates:
$$\int_0^{\frac{\pi}{4}} \int_0^{\sec \theta} \left\{\frac{\cos\left(\theta - \frac{\pi}{4}\right)}{\cos\left(\theta + \frac{\pi}{4}\right)}\right\}r\:dr\:d\theta = \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\{\tan\theta\}\csc^2\theta\:d\theta$$
Now let $u = \cot\theta$ to get
$$\frac{1}{2}\int_0^1 \left\{\frac{1}{u}\right\}\:du = \frac{1-\gamma}{2}$$
which is a somewhat fairly known result, depending on who you ask. The top integral should follow similarly, I'll let you try it on your own (aka left to the reader as an exercise!)
$\textbf{EDIT}$: Given that the inside of the curly braces is negative on the upper piece as @EricTowers noted, we can use
$$\{-x\} = 1 - \{x\}$$
for noninteger $x$ to get that the integral of the upper piece equals
$$\iint_{\text{upper}} \left\{\frac{x+y}{x-y}\right\}\:dx\:dy = \iint_{\text{lower}}1 - \left\{\frac{x+y}{x-y}\right\}\:dx\:dy$$
which implies the sum of the two pieces is simply
$$\iint_{\text{triangle}} dA = \frac{1}{2}$$