Let $I := \int_{(0,1)^2}\frac{1}{1-xy}\, d\lambda^2 (x,y)$.
Can someone help me to determine $I$ only buy using the transformations $u=\frac{1}{2} (y+x)$ and $v=\frac{1}{2} (y-x)$? I don't know how to sketch the new integration area in order to find upper and lower limits of the new integral..
Red: $x=0\implies u=v$
Blue: $y=0\implies u=-v$
Green: $x=1\implies u=v+1$
Yellow: $y=1\implies u=-v+1$. $$ {1\over 1-xy}={1\over1-u^2+v^2}\\ {\partial(u,v)\over\partial(x,y)}=0.5\times0.5-0.5\times(-0.5)=0.5 $$ Integral over the gray area $$ \int_{-0.5}^0\int_v^{-v}{0.5\over1-u^2+v^2}du dv $$ Integral over the white area $$ \int^{-0.5}_{-1}\int^{v+1}_{-v+1}{0.5\over1-u^2+v^2}du dv $$