I'm trying to understand this for the first time. I have to check whether $\int_{0}^{\infty} e^{-\alpha x}sin(x)dx$ is uniformly convergente or not. My guess is it is not convergent if $\alpha \in ]0,\infty[$ but I'm not 100% sure if I proved it right, or how to prove it. So what I have done is:
Let's assume it is uniformly convergent. Then there is a $p \in \mathbb{N}$ such that $\forall \alpha \in]0,\infty[$ $$\left\lvert \int_{p}^{\infty} e^{-\alpha x}sin(x)dx \right\lvert < something$$ not sure what number to put in "something" for a contradiction.
And if this is true then the function $$f(\alpha)= \int_{p}^{\infty} e^{-\alpha x}sin(x)dx \hspace{0.1cm} , \alpha \in ]0,\infty[$$ is bounded. I know that $\lim_{\alpha \to 0} f(\alpha)$ doesn't exist. Is this a contradiction? Why does this contradict the fact that $f$ is bounded? (if $\lim_{\alpha \to 0} f(\alpha) = \infty$ I wouldn't have any doubts, but it doesn't - the limit just doesn't exist so I don't know how to justify).
I hope I was clear about my doubt. Thank you!
The integral is uniformly convergent for $\alpha \in [a,\infty)$ where $a > 0$ by the Weierstrass M-test, but not on $(0,\infty)$.
For the first integral, with $\alpha_n = (2n\pi + \pi)^{-1} \in (0,\infty)$ we have
$$\left|\int_{2n\pi}^{2n\pi+\pi} e^{-\alpha_nx_n} \sin x \, dx\right|\geqslant e^{-(2n\pi+\pi) \alpha_n}\int_{2n\pi}^{2n\pi+\pi} \sin x \, dx = 2 e^{-(2n\pi+\pi)\alpha_n}= 2e^{-1}$$
Since the RHS does not converge to $0$ as $n \to \infty$, the Cauchy criterion for uniform convergence is violated.