Improper integral (is it convergent?) (v 2.0) Earlier today I asked about this question:
Improper integral (is it convergent?)
where the integral fortunately seems to be convergent. So we have that given $\alpha\in (-1/2,0)$ there is a $\gamma \in (1,2)$ such that
$$\int_0^1 \int_0^{u} \frac{((1-v)^{\alpha}-(1-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu < \infty.$$
My question now is, if we take a middle value $a\in (0,1)$ is then the following integral also finite?
$$\int_0^1 \int_0^{u} \frac{((a-v)^{\alpha}-(a-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu < \infty.$$
Of course a naiv start would be to split up the integral as follows:
\begin{align*}
\int_0^1 \int_0^{u} \frac{((a-v)^{\alpha}-(a-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu =& \int_0^a \int_0^{u} \frac{((a-v)^{\alpha}-(a-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu\\
&+ \int_a^1 \int_0^{u} \frac{((a-v)^{\alpha}-(a-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu\\
&=: (A) + (B).
\end{align*}
Here, $(A)$ is essentially the same as the one on top and therefore convergent. So the question is equivalent to proving that $(B)$ is either convergent or divergent.
What are your feelings? Should the integral above still be convergent for any values $a\in (0,1)$ and not only when $a=1$? Any impressions? or ideas on how to prove it?
Thanks a lot guys!

Assuming the first integrate $(a=1)$ is convergent, the second integrate is not such different from the first type. By a simple variable conversion:
$$a-u=1-x$$
$$a-v=1-y$$
The second integral will convert to:
$$ \int_a^{1-a}  \int_{1-a}^x \frac{((1-x)^\alpha-(1-y)^\alpha)^2}{(y-x)^\gamma} dy dx$$ 
which is as same as the first one except for boundaries. If critical points would not make problem in the first integral, they must not make problem in the second one too.
