In proving
$ \Gamma(z) \Gamma(1-z)= \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z} $
appears a $\int_{0}^{\infty}\int_{0}^{\infty}s^{-z}t^{z-1}e^{-(s+t)}\,ds\, dt \qquad (0<\Re z <1)$ where we substitute the value $u=s+t $ and $v=\frac{t}{s}$ so $s= \frac{u}{v+1}$ and $t = \frac{uv}{v+1}$ but sustituing it arrive to
$ Det J(u,v) = 1 \cdot \frac{1}{s} + 1 \cdot \frac{t}{s^2} = \frac{t+s}{s^2}$ which leads to the Formula
$\int_{0}^{\infty}\int_{0}^{\infty}\frac{(v+1)^{3}}{u^2}v^{z-1}e^{-(u)}\,du\, dv \qquad (0<\Re z <1)$
that is different from
$\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(v+1)}v^{z-1}e^{-(u)}\,du\, dv \qquad (0<\Re z <1) \tag {(result)} $
See the thread Detailed explanation of the Γ reflection formula understandable by an AP Calculus student for more explanation.
