<s>[INST] Proving that $\gamma = \int_{0}^{1} \!\!\int_{0}^{1} \!\frac{x - 1}{(1 - x y) \log(x y)} \, \mathrm{d}{x} \, \mathrm{d}{y} $. In 2005, J. Sondow found a surprising formula for the Euler-Mascheroni constant $ \gamma $. The formula is
$$
  \gamma
= \int_{0}^{1} \int_{0}^{1} \frac{x - 1}{(1 - x y) \log(x y)}
  ~ \mathrm{d}{x} ~ \mathrm{d}{y}.
$$
Now, the definition of $ \gamma $ is
$$
\gamma
\stackrel{\text{def}}{=}
\lim_{n \to \infty} \left[ \sum_{k = 1}^{n} \frac{1}{k} - \log(n) \right].
$$
I have tried using the geometric series
$$
\frac{1}{1 - x y} = \sum_{n = 0}^{\infty} x^{n} y^{n}
$$
to obtain a proof, but it would not work. Thanks for any help.
 [/INST] We want use the transformation formula.
\begin{align*}
&U:= \left\lbrace (t,r) \mid t \in (0,1), r \in (t,1) \right\rbrace , V:= (0,1)^2,
\\ &f: V \rightarrow \mathbb{R}, \left( \begin{array}{r}
x\\
y\\
\end{array} \right) \mapsto \frac{x - 1}{(1-xy) \cdot ln(xy)},
\\ &\varphi: U \rightarrow V, \left( \begin{array}{r}
t\\
r\\
\end{array}\right) \mapsto \left( \begin{array}{r}
t/r\\
r\\
\end{array} \right), 
\end{align*}
\begin{align*}
\varphi ^{-1}: V \rightarrow U, \left( \begin{array}{r}
t\\
r\\
\end{array}\right) \mapsto \left( \begin{array}{r}
t \cdot r\\
r\\
\end{array} \right)   
\end{align*}
\begin{align*}
D\varphi = \left( \begin{array}{rr}
1/r & -t/r^2\\
0 & 1\\
\end{array}\right),
Det(D\varphi) = \frac{1}{r}. 
\end{align*}
So we can use the transformation. 
\begin{align*}
\int_0^1 \int_0^1 \frac{x-1}{(1-xy)\cdot ln(xy)} dx \, dy &= \int_0^1 \int_t^1 \frac{(t/r) - 1}{(1-t) \cdot ln(t)} \cdot \frac{1}{r} \, dr \, dt 
\\ &= \int_0^1 \int_t^1 \left( \frac{t}{(1-t) \cdot ln(t)} \cdot \frac{1}{r^2} - \frac{1}{r} \cdot \frac{1}{(1-t) \cdot ln(t)} \right) \, dr \, dt 
\\ &= \int_0^1 \left[ -\frac{1}{r} \cdot \frac{t}{(1-t) \cdot ln(t)} - \frac{ln(r)}{(1-t)\cdot ln(t)} \right]_t^1 \, dt 
\\ &= \int_0^1 \left( \frac{1-t}{(1-t)\cdot ln(t)} + \frac{ln(t)}{(1-t) \cdot ln(t)} \right) \, dt 
\\ &= \int_0^1 \left( \frac{1}{ln(t)} + \frac{1}{1-t} \right) \, dt
\end{align*}
The last integral represented $\gamma$.
</s>