I would like to ask for some help (hints, ideas) in solving the following problem:

Given integer $n>0$ and real $\alpha>0,\beta>1$ we want to show, that

if we define for any $x\in\mathbb{R}^{n}$ the function $$ g(\alpha,x) = \prod\limits_{i=0}^{n-1}(1+\alpha i)\cdot \left(\sum\limits_{i=1}^{n}x_i^{\alpha}\right)^{-\frac{1}{\alpha}-n} \cdot \left(\prod\limits_{i=1}^{n}x_i \right)^{\alpha-1} $$ and the domain $$ A = \{x>{0}:\ \sum\limits_{i} x_i^{\frac{1}{\beta}}>1 \} $$ then the integral given by $$ \int\limits_{A} g(\alpha,x) \mbox{d}x $$ is strictly monotonic with respect to $\alpha$. It could be shown that limits at $0$ and $\infty$ are equal to $n$ and $n^{\beta}$, respectively. It has been proved, that this expression is monotonic. Argument uses representation of $g$ as a limit of certain sequence of functions and cannot prove strict monotonicity.

I tried the following approach which fails: if we show that $g(\alpha,x)$ is analytic with respect to $\alpha$ then, according to the standard procedure (Leibnitz Integral Rule), we prove that integral is analytic. In fact, we easily get for $\alpha>1$ $$|g(\alpha,x)| \leqslant n^{n} \alpha^{n} \\|x\||_{\alpha}^{-n-1} $$ hence $g(\alpha,x)$ is locally uniformly integrable on $A$ with respect to $\alpha$.

However, the expression $ \left(\sum\limits_{i=1}^{n}x_i^{\alpha}\right)^{-\frac{1}{\alpha}-n}$ seems to be non-analytic (with respect to any domain in $\mathbb{C}$ extending positive real line). I wish I knew what other methods could be applied to such kind of problems.