Does this improper integral converge uniformly on $[0,A]$? Does the improper integral$$\int_0^\infty e^{-u^2t}\sin tdu$$ converge uniformly for $t\in[0,A]\quad (A>0)$?
I know that the improper integral converge uniformly for $t\in[a,A]$ where $0<a<A$.Does the improper integral still converge uniformly for $t\in[0,A]$?
 A: Denoting $F(t) = \int_0^\infty e^{-u^2t} \sin t \, du $, you already know that convergence of the improper integral to $F(t)$ is uniform for $t \in [a,A]$ where $a > 0$. The next question should be is the convergence uniform for $t \in (0,A]$ rather than for $t \in [0,A]$, as $F(0)$ is not defined. 
(It can be shown that $F(t) \to 0$ as $t \to 0+$, as the integral has the closed form $F(t) = \frac{\sqrt{\pi}}{2}\frac{\sin t}{\sqrt{t}}$. By defining $F(0) = 0$, the question of uniform convergence for $t$ in the closed interval $[0,A]$ is then relevant and with the same answer.)
Convergence is uniform for $t \in (0,A].$
Note that by making the change of variables $x = u \sqrt{t}$, we have
$$\int_c^\infty e^{-u^2t} \sin t \, du = \frac{\sin t}{\sqrt{t}}\int_{c \sqrt{t}}^\infty e^{-x^2} \, dx = \sqrt{t}\frac{\sin t}{t}\frac{\sqrt{\pi}}{2} \text{erfc}(c\sqrt{t}),$$
where $\text{erfc}$ is the complementary error function.
Thus,
$$\left|\int_c^\infty e^{-u^2t} \sin t \, du \right| = \frac{\sqrt{\pi}}{2}\left|\frac{\sin t}{t} \right| \sqrt{t} \,\text{erfc} (c\sqrt{t}) \leqslant \sqrt{\pi} \sqrt{t} \,e^{-c^2t},$$
where we have used $\left|\frac{\sin t }{t}\right| \leqslant 1$ and the 
Chernoff bound $\text{erfc}(x) \leqslant 2 e^{-x^2}.$
Since $\sqrt{t}e^{-c^2t}$ has a maximum value of $\frac{e^{-1/2}}{\sqrt{2}c}$ at $t = \frac{1}{2c^2}$, we have
$$\lim_{c \to \infty} \sup_{t \in (0,A]}\left|\int_c^\infty e^{-u^2t} \sin t \, du \right| \leqslant \lim_{c \to \infty} \sqrt{\frac{\pi}{2}}e^{-1/2}\frac{1}{c}= 0,$$
and the convergence is uniform for $t\in (0,A].$
