The uniform convergence of $\int_0^\infty e^{-\alpha x}\sin x\,dx$ for $\alpha$ I want to prove that the following improper integral is uniformly convergent with respect to $\alpha\in(0,\infty)$:
$$\int_0^\infty e^{-\alpha x}\sin x\,dx.$$
In other words, I need to show that
$$\forall\epsilon>0, \exists R>0 : \forall t,\alpha\in(0,\infty) \left[t>R\Rightarrow \left|\int_t^\infty e^{-\alpha x}\sin x\,dx\right|<\epsilon\right].$$
The case $\alpha\in[1,\infty)$ is easy since $e^{-\alpha x}\le e^{-x}$ for $x\in(0,\infty)$.
But I could not prove the case $\alpha\in(0,1)$.
Thank you in advance.
 A: The improper integral fails to converge uniformly for $\alpha \in (0,\infty)$ and the problem arises because the upper limit is $+\infty$ and $\alpha$ can be arbitrarily close to $0$.
For a general approach, note that for uniform convergence we require that for any $\epsilon > 0$ there exists $C > 0$ such that for every $c_2 > c_1 > C$ and for every $\alpha \in (0,\infty)$ we have
$$\tag{*}\left|\int_{c_1}^{c_2} e^{-\alpha x} \sin x \, dx\right| < \epsilon$$
Taking $c_1 = 2k\pi$ and $c_2 = (2k+1)\pi$ where $k$ is a positive integer, we have
$$\begin{align} \left|\int_{c_1}^{c_2} e^{-\alpha x} \sin x \, dx\right| &= \int_{2k \pi}^{(2k+1)\pi} e^{-\alpha x} \sin x \, dx \\ &\geqslant  e^{-\alpha(2k+1)\pi}\int_{2k \pi}^{(2k+1)\pi}  \sin x \, dx \\ &= 2e^{-\alpha(2k+1)\pi}\end{align} $$
For any $C > 0$ we can choose a $k$ such that $2k\pi > C$. Choosing $\alpha = [(2k+1) \pi]^{-1}$ we have 
$$\left|\int_{c_1}^{c_2} e^{-\alpha x} \sin x \, dx\right| \geqslant 2e^{-1}$$
and the condition (*) cannot be satisfied for any $\epsilon > 0$.
A: This integral converges for all $\alpha\in (0, \infty)$, but it does not converge uniformly for $\alpha\in (0, \infty)$. Namely, $$\int_t^{\infty} e^{-\alpha x}\sin(x)\,\mathrm{d}x = \frac{e^{-\alpha t}(\cos(t)+\alpha\sin(t))}{1+\alpha^2}$$ which we can find via either integration by parts or Euler's formula. For all $t\in \mathbb{R}$, as $\alpha$ decreases to $0$, the right-hand side approaches $\cos(t)$. For any $T > 0$ and $0 < \epsilon < 1$, we can find a $t\geq T$ such that $\lvert \cos(t)\rvert > \epsilon$, which implies that we can find an $\alpha\in (0, \infty)$ such that $\left\lvert \int_t^{\infty} e^{-\alpha x}\sin(x)\,\mathrm{d}x\right\rvert\geq \epsilon$. This contradicts the uniform convergence condition.
A: If $0 < \alpha <1$, substitute $u = \alpha x$ into your integral. You now have
$$\frac{1}{\alpha}\int_0^\infty e^{-u} \sin(u/\alpha) du$$
which converges by the same point you mentioned.
