Need help with a difficult integral This is  what I have: $$\int_y^1 \frac{\sin\pi z}{z(z-2)}.$$
As you can see this is one part of a triple integral which means it should be simplified for next steps but I don't know how.
UPDATE
The triple one:
$$\int_0^1dx\int_0^{1-x}dy\int_y^1 \frac{\sin\pi z}{z(z-2)}.$$
 A: The direct method should work (using $\;\sin(\pi (z-2))=\sin(\pi z)$) :
\begin{align}
\int_y^1 \frac{\sin(\pi z)}{z(z-2)}dz&=\frac 12\left[\int_y^1\frac{\sin(\pi (z-2))}{z-2}dz-\int_y^1\frac{\sin(\pi z)}{z}dz\right]\\
&=\frac 12\left[\int_{y-2}^{-1}\frac{\sin(\pi u)}{u}du-\int_y^1\frac{\sin(\pi z)}{z}dz\right]\\
&=\frac 12\left[\operatorname{Si}(-\pi)-\operatorname{Si}(\pi (y-2))-\operatorname{Si}(\pi)+\operatorname{Si}(\pi y)\right]\\
&=\frac 12\left[\operatorname{Si}(\pi y)-\operatorname{Si}(\pi (y-2))\right]-\operatorname{Si}(\pi)\\
\end{align}
(from the definition of the sine integral $\;\displaystyle \operatorname{Si}(x):=\int_0^x\frac{\sin(t)}tdt\;$)
Integration by parts allows to obtain : 
$$\int \operatorname{Si}(t)\,dt=[t\,\operatorname{Si}(t)]-\int t\,\operatorname{sinc}(t)\,dt=[t\,\operatorname{Si}(t)+\cos(t)]$$
and :
$$\int t\,\operatorname{Si}(t)\,dt=\left[\frac{t^2}2\,\operatorname{Si}(t)\right]-\int \frac{t^2}2\,\operatorname{sinc}(t)\,dt=\frac 12\left[t^2\,\operatorname{Si}(t)-\sin(t)+t\,\cos(t)\right]$$
So that the triple integral may be done.
I'll let you finish,
