Why is $\lim_{s \to 0} \int_{0}^{\infty} \frac{\sin(t)}{t}\cdot e^{-st} dt =\int_{0}^{\infty} \frac{\sin(t)}{t} dt $ legitimate?

Question

As part of a proof of the value Dirichlet's Integral using only real analysis methods I need to justify the following: $$\lim_{s \to 0} \int_{0}^{\infty} \frac{\sin(t)}{t}\cdot e^{-st} dt = \int_{0}^{\infty} \frac{\sin(t)}{t}\cdot [\lim_{s \to 0} e^{-st}] dt = \int_{0}^{\infty} \frac{\sin(t)}{t} dt $$ Specifically I want to justify rigorously why I can bring the $\lim$ process into the integral (or out of). The way I see it, I need to show uniform integrability of $\lim_{s \to 0} \int_{0}^{\infty} \frac{\sin(t)}{t}\cdot e^{-st} dt$ independtent of $s$ using integration by parts. I can't figure out the technical details though.

Answer 1

The integral

$$F(s) = \int_0^\infty f(t,s) \, dt= \int_0^\infty \frac{ \sin t}{t} e^{-st} \, dt$$

converges uniformly for $s \geqslant 0$. This follows from the Dirichlet test since $\left|\int_0^x \sin t \, dt\right|$ is uniformly bounded and $e^{-st}/t $ is decreasing and uniformly convergent to $0$ as $t \to \infty$.

Given uniform convergence to $F(s)$ and continuity of $(t,s) \mapsto f(t,s)$ we can apply a basic theorem found in standard real analysis textbooks (Rudin, Apostol, Bartle, etc.) proving $F$ to be continuous, so

$$\lim_{s \to 0}F(s) = F(0) = \int_0^\infty \frac{\sin t}{t} \, dt $$