An interesting representation of the Euler-Mascheroni constant
$$ \gamma~=~ \lim \limits\_{n\to \infty} \sum \limits\_{k,s=1}^n \frac{s-k}{k\left( s\,n +k\right)},\label{1}\tag{$\*$}$$
can be proved using the relatively recent identity of Macys (Mat. Zametki, 94/5, 695–701, 2013), but it also can be reformulated to
almost mimic a Riemann sum over the unit square $[0,1]^2$
$$ \gamma~=~ \lim \limits\_{n\to \infty} \sum \limits\_{x,y=1/n\,(step: 1/n)}^1 \frac{x-y}{y\,\left( x +y/n\right)} \,\left(\frac1n\right)^2,$$
and there are known similar representations as double integral like
$$ \gamma~=~ \int \limits\_0^1 \int \limits\_0^1 \frac{1-x}{(1-x\, y ) (-\log x\,y)}\,dx\,dy. $$
Is there a valid way to prove \eqref{1} by converting it into an integral representation ?



An Euler-Mascheroni double sum
$\newcommand\ga\gamma$As usual, let $H\_n$ denote the $n$th harmonic number. Note that
$$\frac{s-k}{k (s n+k)}=\frac1n\frac1k-\frac{n+1}n\frac1{n s+k}$$
and hence
$$
\begin{aligned}
&\sum\_{k,s=1}^n\frac{s-k}{k (s n+k)} \\
&=\frac1n\,\sum\_{k,s=1}^n\frac1k-\frac{n+1}n\,\sum\_{k,s=1}^n\frac1{n s+k} \\ 
&=H\_n-\frac{n+1}n\,\sum\_{s=1}^n\sum\_{k=1}^n\frac1{n s+k} \\ 
&=H\_n-\frac{n+1}n\,\sum\_{s=1}^n (H\_{n(s+1)}-H\_{ns}) \\ 
&=H\_n-\frac{n+1}n\,(H\_{n(n+1)}-H\_n) \\ 
&=\frac{2n+1}n\,H\_n-\frac{n+1}n\,H\_{n(n+1)}\\ 
&=\frac{2n+1}n\,(\ga+\ln n+o(1)) \\ 
&-\frac{n+1}n\,(\ga+\ln(n(n+1))+o(1)) \\ 
&=\ga+o(1)
\end{aligned}$$
(as $n\to\infty$), which proves your formula $(\*)$.
Of course, $H\_n$ can be represented as an integral; for instance, we have [Euler's representation](https://en.wikipedia.org/wiki/Harmonic_number#Calculation)
$$H\_n=\int \_{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx.$$




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On the other hand, apparently, your Riemann-sum argument cannot work -- because the boundary effect here is nonnegligible. In particular, summing over the set $\{2,\dots,n\}\times\{2,\dots,n\}$ rather than over the set $\{1,\dots,n\}\times\{1,\dots,n\}$, we get
$$\lim\_{n\to\infty}\sum\_{k,s=\color{red}{\mathbf{2}}}^n\frac{s-k}{k (s n+k)}=\ga-1+\ln2\ne\ga.$$


