= Solution
{c}
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 $e^{-\alpha x^\beta}\leq x^{-2}$ 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, $\alpha\geq a:=\alpha_0/2>0$ and $\beta\geq b:=\beta_0/2>0$. For every parameter pair in this rectangle,
$$
0\leq e^{-\alpha x^\beta}\leq H(x):=\begin{cases}1,&0<x<1,\\e^{-a x^b},&x\geq1.\end{cases}
$$
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>:
$$
\boxed{g(\alpha,\beta)=\frac{\Gamma(1/\beta)}{\beta\alpha^{1/\beta}}.}
$$
The domination argument establishes <continuity> directly, without needing a separate theorem about parameter dependence of the <gamma function>.
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