If $f(x,t)$ satisfies $|f(x,t)| < g(t)$ for all $x \in A$ and $\int_a^\infty g(t)dt$ converges, then $\int_a^\infty f(x,t)dt$ converges uniformly on $A$.
I understand that uniform convergence for improper integrals of this type means that assuming $F(x) = \int_a^\infty f(x,t)dt$ exists (converges) for all $x \in A$, and given an $x \in A$ and $\epsilon > 0$, we can find an $M > a$ s.t.
$$ \bigg{|} F(x) - \int_a^d f(x,t)dt \bigg{|} < \epsilon$$
for all $ d\geq M$. I've been able to show that the improper integral $\int_a^\infty f(x,t)dt$ converges absolutely, but I'm not what this says, if anything, about uniform convergence.
Since the improper integral of $g$ is convergent, for any $\epsilon > 0$ there exists $C > a$ (independent of $x$) such that for all $c_2 > c_1 > C$ we have
$$\left| \int_{c_1}^{c_2} f(x,t) \, dt \right| \leqslant \int_{c_1}^{c_2} |f(x,t)| \, dt \leqslant \int_{c_1}^{c_2} g(t) \, dt < \epsilon$$
As this inequality holds for all $x \in A$, convergence is uniform by the Cauchy criterion for uniform convergence of improper integrals. See Theorem 2 here for a proof of this criterion.