Check the uniform convergence of parametric integral While $0< \alpha <+ \infty$, prove if the parametric integral is uniform convergent on $\alpha$'s domain:
$$\int^{+ \infty}_{0} e^{- \alpha x} \sin \beta x dx$$
$\beta$ is nonzero constant.
 A: The integral is uniformly convergent for $\alpha \in[\delta,\infty)$ for each $\delta >0$.
This follows from the Weierstrass Test since
$$|e^{-\alpha x}\sin{\beta x}| \leq e^{-\delta x}$$
for $\alpha \geq \delta$ and the integral of the RHS converges:
$$\int_{0}^{\infty}e^{-\delta x}dx= \frac1{\delta}$$
However, it is not uniformly convergent for $\alpha \in (0,\infty)$.
To prove non-uniform convergence we can show there are sequences $\alpha_n \rightarrow 0$ and $b_n < c_n \rightarrow \infty$ such that
$$\int_{b_n}^{c_n}e^{-\alpha_n x}\sin{\beta x} dx \geq \frac1{4 \beta}.$$ Assuming WLOG $\beta > 0$, choose $0 < b < c$  such that
$$\int_{b}^{c} \sin{\beta x} dx = \frac1{\beta}\int_{\beta b}^{\beta c} \sin{x} dx \geqslant \frac{1}{2\beta}.$$
Let $b_n = 2\pi n + b$ and $c_n = 2\pi n + c$. Then
$$\int_{b_n}^{c_n}e^{-\alpha_n x}\sin{\beta x} dx \geq e^{-\alpha_n c_n}\int_{b_n}^{c_n}\sin{\beta x} dx \geq \frac1{2 \beta}e^{-\alpha_n c_n}.$$
Now let $\alpha_n$ be a sequence that converges to $0$ and satisfies 
$$\alpha_n < \frac{\log(2)}{2\pi n  + c}.$$
Then
$$\int_{b_n}^{c_n}e^{-\alpha_n x}\sin{\beta x} dx \geq\frac1{2}e^{-\alpha_n c_n} > \frac1{4 \beta}.$$
Note, also, that for $\alpha = 0$, the integral diverges (oscillates).
