Does $\int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x}dx$ converge uniformly

Question

Does $$\int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x}dx \hspace{0.1cm}, \alpha \in ]0,\infty[$$ converge uniformly?

Using the Dirichlet test:

1. $\int_0^\infty \frac{\sin(x)}{x}dx = \pi/2$
2. $e^{-\alpha x}$ is decreasing, bounded and going to $0$.

So it converges uniformly.

Is this ok? Or does it only converge uniformly in $]k,\infty[$ with $k>0$ ?

Answer 1

Hint to use the Dirichlet test:

We have $\int_0^c \sin x \, dx$ bounded for all $c > 0$ and independent of $\alpha$ and $\frac{e^{-\alpha x}}{x}$ is monotonically decreasing in $x$ and uniformly convergent to $0$ as $x \to \infty$ for all $\alpha \in [0,\infty)$.