I'm trying to calculate a closed form of this double integral
$$\int_0^1\int_0^1(1-y)^a\,x^{b\,y-1}(x-1/2)^n\,dx\,dy \tag{1}$$
with $a>-1$, $b>0$ and $n=0,1,2,\dots$
If we apply the Newton binomial to $(z-1/2)^n$ our problem reduces to calculate integrals of the form
$$\int_0^1\int_0^1(1-y)^a\,x^{b\,y-1+n}\,dx\,dy$$
Those integrals are easy to calculate except the case $n=0$, that is,
$$\int_0^1\int_0^1(1-y)^a\,x^{b\,y-1}\,dx\,dy$$
I've checked $(1)$ with Wolfram and it seems to be finite. I've also checked in several books of definite integrals but I've had no luck.
Any help will be welcome.
I get $$\begin{align}\int_0^1\int_0^1(1-y)^ax^{by-1}dx\,dy&=\int_0^1(1-y)^a\left.\frac{x^{by}}{by}\right|_0^1dy=\int_0^1\frac{(1-y)^a}{by}dy\\ &>\int_0^{1/2}\frac{dy}{2^aby}=\left.\frac{\ln y}{2^ab}\right|_0^{1/2}=\infty\end{align}$$ Indeed WA agrees with one sample integral.