The uniform convergence of $\int_0^\infty e^{-\alpha x}\sin x\,dx$ for $\alpha$

Question

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.

Answer 1

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 $\epsilon > 0$.

Answer 2

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.

Answer 3

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.