An interesting representation of the EulerMascheroni constant $$ \gamma~=~ \lim \limits_{n\to \infty} \sum \limits_{k,s=1}^n \frac{sk}{k\left( s\,n +k\right)},\label{1}\tag{$*$}$$ 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 $[0,1]^2$ $$ \gamma~=~ \lim \limits_{n\to \infty} \sum \limits_{x,y=1/n\,(step: 1/n)}^1 \frac{xy}{y\,\left( x +y/n\right)} \,\left(\frac1n\right)^2,$$ and there are known similar representations as double integral like $$ \gamma~=~ \int \limits_0^1 \int \limits_0^1 \frac{1x}{(1x\, y ) (\log x\,y)}\,dx\,dy. $$ Is there a valid way to prove \eqref{1} by converting it into an integral representation ?
1 Answer
$\newcommand\ga\gamma$As usual, let $H_n$ denote the $n$th harmonic number. Note that $$\frac{sk}{k (s n+k)}=\frac1n\frac1k\frac{n+1}n\frac1{n s+k}$$ and hence $$ \begin{aligned} &\sum_{k,s=1}^n\frac{sk}{k (s n+k)} \\ &=\frac1n\,\sum_{k,s=1}^n\frac1k\frac{n+1}n\,\sum_{k,s=1}^n\frac1{n s+k} \\ &=H_n\frac{n+1}n\,\sum_{s=1}^n\sum_{k=1}^n\frac1{n s+k} \\ &=H_n\frac{n+1}n\,\sum_{s=1}^n (H_{n(s+1)}H_{ns}) \\ &=H_n\frac{n+1}n\,(H_{n(n+1)}H_n) \\ &=\frac{2n+1}n\,H_n\frac{n+1}n\,H_{n(n+1)}\\ &=\frac{2n+1}n\,(\ga+\ln n+o(1)) \\ &\frac{n+1}n\,(\ga+\ln(n(n+1))+o(1)) \\ &=\ga+o(1) \end{aligned}$$ (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 $$H_n=\int _{0}^{1}{\frac {1x^{n}}{1x}}\,dx.$$
On the other hand, apparently, your Riemannsum 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 $$\lim_{n\to\infty}\sum_{k,s=\color{red}{\mathbf{2}}}^n\frac{sk}{k (s n+k)}=\ga1+\ln2\ne\ga.$$

$\begingroup$ Thanks. Your proof reconstructs that the formula is based on the idea of Macys that $2\,H_nH_{n^2} \to \gamma$. I like your argument that the nonintegrable divergence at the zero boundaries destroys any integral proof route. $\endgroup$ Sep 8, 2021 at 22:12