TITLE: Evaluating $I(z,s)=\int_0^1\int_0^1\left(1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z)y)}\right)^{s-2}\,\mathrm dx\mathrm dy$
QUESTION [6 upvotes]: I came across the following double integral in a statistics problem:
For $z>0$
$$
I(z,s)=\int_0^1\int_0^1\left(1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z)y)}\right)^{s-2}\,\mathrm dx\mathrm dy.
$$
All I was interested in was the values of $s$ for which $I$ converges $(s>0)$. That said, the integral looked interesting enough to try and evaluate.  I was successful in evaluating the special case $z=1$ (see below) but could not solve for the general solution. Any ideas on how to solve it?

Case for $z=1$:
We have
$$
I(1,s)=\int_0^1\int_0^1\left(1-(1-x)(1-y)\right)^{s-2}\,\mathrm dx\mathrm dy.
$$
Substituting $(u,v)=(1-(1-x)(1-y),x)$, the resulting integral in $v$ is easily evaluated in terms of the natural logarithm; hence,
$$
I(1,s)=-\int_0^1 u^{s-2}\log(1-u)\,\mathrm du.
$$
Observe that $-u^{s-2}\log(1-u)\sim u^{s-1}+\mathcal O(u^s)$ as $u\nearrow 0$, which tells us the integral should converge for $s>0$. Using the series expansion for the logarithm we then integrate termwise to find
$$
I(1,s)=\sum_{k=1}^\infty\frac{1}{k(k+s-1)}=\frac{H_{s-1}}{s-1},
$$
where $H_{s-1}=\psi(s)+\gamma$ is the generalized harmonic number, $\psi(z)=\partial_z\log\Gamma(z)$ is the digamma function and $\gamma=0.577\dots$ is the Euler–Mascheroni constant.

REPLY [3 votes]: This is meant to supplement the solution derived by @jwimberley. The solution derived was (after some algebraic manipulations):
$$
I(z,s)=\sum_{k=0}^{s-2}\frac{(2-s)_k}{k!}\frac{F^2(1,k;k+2;1-z)}{(k+1)^2},
$$
which held for $z>0$ and $s\in\Bbb N$ with $s\geq 2$. I would like to show that this solution in fact holds for all $s\in\Bbb C: \Re s>0$. First note that the solution above is equivalent to
$$
I(z,s)=1+\sum_{k=1}^\infty\frac{(2-s)_k}{(k+1)^2k!}F^2(1,k;k+2;1-z),
$$
since the first term is always one and $s\in\Bbb N\land s\geq 2\implies (2-s)_k=0$ for $k=s-1,s,s+1,\dots$
To show convergence for other values of $s$ we will use the integral representation of the hypergeometric function (DLMF $15.6.1$) to write
$$
F\left({1,k \atop k+2};1-z\right)=\int_0^1\frac{1}{1-(1-z) t} \frac{t^{k-1}(1-t)}{\operatorname{B}(k,2)}\,\mathrm dt,\quad k\in\Bbb N
$$
which can be interpreted as an expected value w.r.t. the $\operatorname{Beta}(k,2)$ distribution, i.e. $\mathsf E[(1-(1-z) X)^{-1}]$ with $X\sim\operatorname{Beta}(k,2)$. By Jensen's inequality we then have
$$
F^2\left({1,k \atop k+2};1-z\right)\leq \mathsf E[(1-(1-z) X)^{-2}]=\int_0^1\frac{1}{(1-(1-z) t)^2} \frac{t^{k-1}(1-t)}{\operatorname{B}(k,2)}\,\mathrm dt.
$$
Now note for $z>0$ and $t\in[0,1]$ that
$$
\min\{1,z^{-2}\}\leq\frac{1}{(1-(1-z) t)^2}\leq\max\{1,z^{-2}\}.
$$
Consequently,
$$
F^2\left({1,k \atop k+2};1-z\right)\leq\max\{1,z^{-2}\}
$$
and
$$
\tag{1}
I(z,s)\leq1+\max\{1,z^{-2}\}\left(\sum_{k=0}^\infty\frac{(2-s)_k}{(k+1)^2k!}-1\right).
$$
By the properties of the Pochhammer symbol we now write $k+1=(2)_k/(1)_k$. Substituting this result into our inequality for $I$ gives
$$
I(z,s)\leq1+\max\{1,z^{-2}\}\left(\sum_{k=0}^\infty\frac{(1)_k(1)_k(2-s)_k}{(2)_k(2)_kk!}-1\right),
$$
which according to DLMF $\S 16.2(\mathrm{iii})$ is absolutely convergent if $\Re s>0$. Furthermore, if we evaluate the series we can obtain an explicit expression for the upper bound on $I$. Upon inspection of $(1)$ we see that
$$
I(z,s)\leq1+\max\{1,z^{-2}\}\left(I(1,s)-1\right).
$$

Hence, for $z>0$ and $s\in\Bbb C:\Re s>0$:
$$
I(z,s)=\sum_{k=0}^\infty\frac{(2-s)_k}{k!}\frac{F^2(1,k;k+2;1-z)}{(k+1)^2}
$$
and
$$
I(z,s)\leq1+\max\{1,z^{-2}\}\left(\frac{H_{s-1}}{s-1}-1\right).
$$