 Solution

On $(0,1)$, the integrand is bounded by one. For $x\to \infty$, $\alpha x^{\beta }-2\log x\to \infty$, since every positive power grows faster than the natural logarithm. Thus $e^{-\alpha x^{\beta }}\le x^{-2}$ eventually. The comparison test for integrals proves Lebesgue integrability.

To prove joint continuity, fix $({\alpha }_0,{\beta }_0)$ with both coordinates positive. In a sufficiently small parameter rectangle, $\alpha \ge a:={\alpha }_0/2>0$ and $\beta \ge b:={\beta }_0/2>0$. For every parameter pair in this rectangle,The first argument shows that $H$ is integrable. Along any parameter sequence tending to $({\alpha }_0,{\beta }_0)$, the integrands converge pointwise, and the same $H$ dominates their eventual tail. The dominated convergence theorem gives $g(\alpha ,\beta )\to g({\alpha }_0,{\beta }_0)$. The sequential criterion for continuity proves the required joint continuity throughout the positive quadrant.

For comparison, the change of variables $u=\alpha x^{\beta }$ identifies the value with a gamma function:The domination argument establishes continuity directly, without needing a separate theorem about parameter dependence of the gamma function.