What are the rules used for this integral representation? If we have $f(w)= w^{-\frac{\alpha}{\beta}}\displaystyle\int_0^w \frac{z^{\frac{\alpha}{\beta}-1}}{1-z} \mathrm{d} z $, what are the rules used to form $f(w)=\displaystyle\int_0^1 \frac{u^{\frac{\alpha}{\beta}-1}}{1-w u} \mathrm{d} u$?
Thanks a lot.

For fixed $w$, we consider $w^{-\frac{\alpha}{\beta}}\displaystyle\int_0^w \frac{z^{\frac{\alpha}{\beta}-1}}{1-z} \mathrm{d} z$, which you called $f(w)$. 
Let $z=wu$. Hence, if $z=w$, then $u=1$; if $z=0$, then if $z=0$, then $u=1$. Also,  $\mathrm{d} z=w\,\mathrm{d}u$. On the other hand, 
$$\frac{z^{\frac{\alpha}{\beta}-1}}{1-z}=\frac{w^{\frac{\alpha}{\beta}-1}u^{\frac{\alpha}{\beta}-1}}{1-wu}.$$
Now putting all these calculations together, we obtain
$$w^{-\frac{\alpha}{\beta}}\displaystyle\int_0^w \frac{z^{\frac{\alpha}{\beta}-1}}{1-z} \mathrm{d} z=w^{-\frac{\alpha}{\beta}}\int_0^1\frac{w^{\frac{\alpha}{\beta}-1}u^{\frac{\alpha}{\beta}-1}}{1-wu}\cdot w\,\mathrm{d}u=\int_0^1\frac{u^{\frac{\alpha}{\beta}-1}}{1-wu} \mathrm{d}u,$$
as required.
