If $|h(x,t)|\leq g(t)$ then the integral $f(x) = \int_a^{\infty} h(x,t)dt$ is uniformly convergent.
Solution 1:
Define the sequence of functions
$$f_N (x) := \int_a^N h(x,t) dt.$$
Let $\epsilon >0$.
\begin{align} \left|f(x) - f_N(x) \right| \leq & \int_N^\infty \left|h(x,t) \right| dt \\ &\leq \int_N^\infty \left|g(t) \right| dt \end{align} Choose $N>0$ large enough such that $\int_N^\infty \left|g(t) \right| dt < \epsilon$, then we have
$$\left|f(x) - f_N(x) \right| \leq \epsilon, \,\,\, \forall x \in I $$
Thus $$\int_a^N h(x,t) dt=: f_N(x) \underset{\text{uniformly}}{\to} f(x)=\int_a^\infty h(x,t) dt $$ The convergence is uniform because the dependence is just for $t$.