A nice double integral on the square $[0,1]$x$[0,1]$ It is recalled that the fractional part of a real $x$, noted {$x$}, is defined by
{$x$} $=x-\lfloor x\rfloor $ where $\lfloor x\rfloor $ is the floor function (largest integer not greater than $x$, many times noted $[x]$).
I don’t remember where I saw the double integral below but I have never forgot it gives us a very nice equality: it is equal to a linear expression $a\gamma+b$
    of the important Euler’s Constant $\gamma$ with very simple rational coefficients $a$ and $b$.
 It is well known that the constant $\gamma$ is defined by 
$$\gamma=\lim_{n\rightarrow \infty}(\sum_{k=1}^n \frac{1}{k}-\ln n)$$ being perhaps the most celebrated limit of the indeterminate form $\infty-\infty$ 
So we know $$\int_0^1\int_0^1\left\{\frac{x}{y}\right\}dxdy=a\gamma +b$$
Determine the values of rational $a$ and $b$.

$$\int_0^1\int_0^1\{\frac{x}{y}\}dxdy \\
= \lim_{N\to\infty} \sum_{n=1}^{N}\int_{\frac{1}{n+1}}^{\frac{1}{n}}((\sum_{k=0}^{[\frac{1}{y}]-1}\int_{ky}^{(k+1)y} (\frac{x}{y}-[\frac{x}{y}]) dx) + \int_{ny}^1 (\frac{x}{y}-[\frac{x}{y}])dx) dy \\
= \lim_{N\to\infty} \sum_{n=1}^{N} \frac{1}{2}\ln(\frac{n+1}{n})-\int_{\frac{1}{n+1}}^{\frac{1}{n}}(\sum_{k=0}^{n-1}\int_{ky}^{(k+1)y} [\frac{x}{y}] dx + \int_{ny}^1 [\frac{x}{y}]dx) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \int_{\frac{1}{n+1}}^{\frac{1}{n}}(\sum_{k=0}^{n-1}\int_{ky}^{(k+1)y} k dx + \int_{ny}^1 ndx) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \int_{\frac{1}{n+1}}^{\frac{1}{n}}(\sum_{k=0}^{n-1}ky + n(1-ny)) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \int_{\frac{1}{n+1}}^{\frac{1}{n}}(y\frac{n(n-1)}{2} + n(1-ny)) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \int_{\frac{1}{n+1}}^{\frac{1}{n}}(-y\frac{n(n+1)}{2} + n) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \frac{2n-1}{4n(n+1)}\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \frac{1}{4} \sum_{n=1}^{N} (-\frac{1}{n}+\frac{3}{n+1})\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)-\frac{1}{2}\sum_1^{N+1}\frac{1}{n} + \frac{3}{4}-\frac{1}{4(N+1)}\\
=-\frac{1}{2}\gamma+\frac{3}{4}$$
