I am working through a paper I posted to arXiv and am looking to condense my results as well as seek alternative proofs for some of my intermediate results. In particular, I am looking for a shorter proof that gets me to the result of Theorem 1 below. This Theorem was used to justify convergence of a more complicated double integral in my paper. Is Lemma 1 needed at all? Is there a much simpler line of reasoning to get to Theorem 1?
Lemma 1.
For any $n\in\Bbb N_0$, $$ I_n=\int_0^1\int_0^1(-1)^n\log^n(1-(1-x)(1-y))\,(1-(1-x)(1-y))^{s-2}\,\mathrm dx\mathrm dy, $$ converges if $s>0$.
Proof.
We will first evaluate $I_0$. Perform the change of variables $(t,v)=((1-x)(1-y),x)$ and integrate over $v$ yielding $$ I_n=\int_0^1(-1)^n\log^n(1-t)(-\log t) (1-t)^{s-2}\,\mathrm dt. $$ Substituting $n=0$, we again change variables via $x=1-t$ and then integrate by parts with $u=-\log(1-x)$ and $\mathrm dv=x^{s-2}\,\mathrm dx$ to find $$ I_0=\frac{1}{s-1}\int_0^1\frac{1-x^{s-1}}{1-x}\,\mathrm dx+\log(1-x)\frac{1-x^{s-1}}{s-1}\bigg|_{x=0}^1. $$ If $s>0$ the limit term vanishes and upon inspection of the integral representation for the harmonic numbers $$ I_0=\frac{H_{s-1}}{s-1}. $$ Now consider the general case $I_n$. Without loss of generality assume $n\geq 1$ and perform integration by parts with $u=(-1)^n\log^n(1-t)$ and $\mathrm dv=-\log t(1-t)^{s-2}\,\mathrm dt$. Expanding the logarithm in $\mathrm dv$ as a power series in $(1-t)$ and integrating termwise we find $$ v=-\sum_{k=0}^\infty\frac{(1-t)^{s+k}}{(s+k)(1+k)}. $$ In this form, it becomes clear that the limit term $uv|_{t=0}^1$ vanishes if $n\geq 1$ so that $I_n=\int_0^1(-v)\,\mathrm du$. Furthermore, we observe for $s>0$: $$ -v=\frac{1}{s}(1-t)^{s-1}\sum_{k=0}^\infty\frac{s(1-t)^{k+1}}{(s+k)(1+k)}\leq \frac{1}{s}(-\log t)(1-t)^{s-1}. $$ Hence, $$ I_n\leq\frac{n}{s}\int_0^1(-1)^{n-1}\log^{n-1}(1-t)(-\log t)(1-t)^{s-2}\,\mathrm du=\frac{n}{s}I_{n-1}. $$ Solving the recurrence relation and calling on the result for $I_0$ we find for $s>0$: $$ I_n\leq\frac{n!}{s^n}\frac{H_{s-1}}{s-1}<\infty, $$
Theorem 1.
Let $z\in\Bbb R^+$, $m,n\in\Bbb N_0$, $a,b\in(-\infty,2]$, $$ f(x,y,z) =1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z) y)}, $$ $$ F(x,y,z,l,s)=(-1)^l(\log\circ f)^l(x,y,z)f^{s-2}(x,y,z), $$ and $$ I=\int_0^1\int_0^1F(x,y,z,n,a)F(x,y,1,m,b)\,\mathrm dx\mathrm dy. $$ Then $I$ converges whenever $a+b>2$.
Proof.
For convenience we will introduce the auxiliary function $$ J(x,y,z)=\frac{z^2}{(z+(1-z)x)^2(z+(1-z)y)^2} $$ as well as the following properties: $$ \begin{array}{*2{>{\displaystyle}l}} (\mathrm{FII}) &\text{$f(\cdot,z)$ is increasing on $z\in\Bbb R^+$ from $f(\cdot,0)=0$ to $f(\cdot,\infty)=1$}.\\[1ex] (\mathrm{J}) &\text{For all $z\in\Bbb R^+$ and $(x,y)\in[0,1]^2$: $0\leq J(x,y,z)\leq z^{2\operatorname{sign}(z-1)}$}. \end{array} $$ Since $n\in\Bbb N_0$ and $a\leq 2$ we are able to deduce from property $(\mathrm{FII})$ that $F(x,y,z,n,a)$ is nonnegative and a decreasing function of $z$ for all $z\in\Bbb R^+$. Denoting $z^\ast=\min\{1,z\}$, it follows for all $z\in\Bbb R^+$ $$ I\leq\int_0^1\int_0^1F(x,y,z^\ast,m+n,a+b-2)\,\mathrm dx\mathrm dy. $$ Performing the change of variables $(1-u,1-v)=((1-x)/(1-(1-z^\ast)x),(1-y),(1-(1-z^\ast)y))$ and then using property $(\mathrm J)$ subsequently gives $$ I\leq\max\{1,z^{-2}\}\int_0^1\int_0^1F(u,v,1,m+n,a+b-2)\,\mathrm du\mathrm dv, $$ which according to the Lemma 1 converges if $a+b>2$. The proof is now complete.