I am new to proving uniform convergence and am trying to show that $$f_{s}(t) = \frac{e^{-st}}{t}$$ is uniformly convergent $\forall$ $s\geq 0$. I am pretty sure that $t\epsilon \mathbb{R}$. I first thought of taking the derivative of the function to find the maximum since t is in an open set, but I remember that since I am showing that $\left |f_{s}(t)\right |<\varepsilon$, and knowing that the value of $f_{s}(t)$ when $t<0$ (the value of the local maximum of $f_{s}(t)$ is also less than $0$) is negative, if one were to take the absolute value, then the local maximum would become a local minimum. Also, there would not be a maximum, since $\lim _{s\rightarrow0} \rightarrow+\infty$. I know that the answer to this question is that $f_{s}(t)$ is uniformly convergent to $0$ as $t\rightarrow+\infty$,$\forall$ $s\geq 0$, but I am stuck at the point where I am at right now. Is there any theorem or something that I can use to help me with my proof?
Edit: This is in the context to use the Dirichlet test to show that an integrand composed of a product is uniformly convergent, so taking that in mind, should it be restricted to only the positive real numbers?
For uniform convergence as $t \to +\infty$ you can consider the behavior when $t$ is positive. For all $s \geqslant 0$ we have
$$0 < \frac{e^{-st}}{t} \leqslant \frac{1}{t}$$
and it is now easy to conclude that convergence is uniform.