Double Integral: solution with hypergeometric function?

[Dirickby]

Homework Statement

Hello, I've recently encountered this double integral
$$\int_0^1 dv \int_0^1 dw \frac{(vw)^n(1-v)^m}{(1-vw)^\alpha} $$

with $\Re(n),\Re(m) \geq 0$ and $\alpha = 1,2,3$.

Homework Equations

I use Table of Integrals, Series and Products by Gradshteyn & Ryzhik as a reference. There I found that
$$\int_0^1 dv \frac{(v)^n(1-v)^m}{(1-vw)^\alpha} = \beta(n+1, m+1)
_2F_1(\alpha ,n+1; n+m+2; w) \qquad\text{(3.197.3)} $$
with the $\beta$-function and $_2F_1$ the ordinary hypergeometric function. But this equation is only valid for $w<1$ (and $\Re(n+1),\Re(m+1) > 0$, which is the case here).

The Attempt at a Solution

If I could use the formula above, I could easily integrate over $v$ and then over $w$, but I suppose it would be wrong as $w=1$ at the upper integration limit. Can I somehow bypass this?
Additionnally if we were to only consider $w=1$, the integral would diverge at least in some cases (e.g. $m=0$). Could my whole integral diverge because of this?

 

[DrDu]

At w=1, the integral from Gradstein diverges logarithmically. But the integral over a logarithmic singularity is finite, so the integral over w should exist.