$\displaystyle

\int_0^1 \int_0^{1-x} \int_y^1 \frac {\sin (z\pi)}{z(2-z)} dz dy dx

$

I first showed the function is integrable,

$\displaystyle

\lim_{ z\to 0} f(x.y.z)=\frac{\pi}{2}

$

and we know z will never be 2

so the function can be integrate.

How do I go on from here?