Study the uniform convergence of the integral $\int_0^\infty\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x, \alpha\in(0,+\infty)$ 
Study the uniform convergence of the integral
$$\int_0^\infty\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x,\quad \alpha\in(0,+\infty)$$

I have tried the Weierstrass criterion, but $\int_0^\infty\frac{1}{\alpha^2+x^2}\mathrm{d}x$ does not converge. Then I tried the Dirichlet criterion, but $\int_0^A \sin \alpha x$ are not bounded uniformly. So I turned to prove it does not converge uniformly. My attempt are as follows
$\forall A>1$, choose $\alpha=\frac{\pi}{4A}$ then $\forall x\in[A, 2A]$
$$
\left|\int_A^{2A}\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x\right|>\frac{\sqrt{2}}{2}\frac{1}{\alpha}(\arctan\frac{2A}{\alpha}-\arctan\frac{A}{\alpha})
$$
but I cannot find a lower bound of RHS, so the method seems do not work.
Is my way wrong? You can also suggest your way. Appreciate any help!
 A: The improper integral
$$I(\alpha) = \int_0^\infty \frac{\sin \alpha x}{\alpha^2 + x^2} \, dx$$
converges uniformly by the Weierstrass-M test on any interval $[A,\infty)$ where $A > 0$ since for all $\alpha > A$,
$$\left|\frac{\sin \alpha x}{\alpha^2 + x^2} \right| \leqslant\frac{1}{A^2 + x^2} \,\,\text{and }\,\, \int_0^\infty\frac{dx}{A^2 + x^2 }< \infty$$
We can also show that convergence is uniform on $(0,A]$ and, hence, on $(0,\infty]$ using the Dirichlet test.
We can rewrite the integral as
$$I(\alpha) = \int_0^\infty \frac{x}{\alpha^2+x^2} \frac{\sin \alpha x}{x} \, dx$$
With the change of variables $t = \alpha x$ we have
$$\int_0^c \frac{\sin \alpha x}{x} \, dx = \int_0^{\alpha c} \frac{\sin t}{t} \, dt = \mathrm{Si}(\alpha c) $$
and the integral (see  sine integral ) is uniformly bounded for all $\alpha,c \in [0,\infty)$.
We also have the uniform convergence $f(x,\alpha) = \frac{x}{\alpha^2+x^2} \leqslant \frac{1}{x} \to 0$ as $x \to \infty$. Furthermore,  $f(x, \alpha)$ is monotonically decreasing with respect to $x$ for all $x \geqslant A$ independent of $\alpha \in (0,A]$. Note that $f(0,\alpha) = 0$, there is a maximum for $x = \alpha$, and then $f(x,\alpha)$ is decreasing to $0$ for all $x > \alpha$. Therefore, the improper integral is uniformly convergent for $\alpha \in (0,A]$.

Note that we only need eventual monotonicity of $f$ (independent of the parameter) to retain applicability of the Dirichlet test.  To see this, suppose $f(x,\alpha) \searrow 0$ uniformly and $f(x,\alpha)$ is decreasing with respect $x$ when $x \geqslant A$.
For all $c_2 > c_1 >A$, by the second mean value theorem for integrals (which requires monotonicity of $f$), there exist $\xi \in (c_1,c_2)$ such that
$$\int_{c_1}^{c_2} f(x,\alpha) g(x,\alpha) \, dx = f(c_1,\alpha) \int_{c_1}^\xi g(x, \alpha) \, dx +  f(c_2,\alpha) \int_{\xi}^{c_2} g(x, \alpha) \, dx, $$
and
$$\left|\int_{c_1}^{c_2} f(x,\alpha) g(x,\alpha) \, dx\right| \leqslant |f(c_1,\alpha)| \left|\int_{c_1}^\xi g(x, \alpha) \, dx \right|+  |f(c_2,\alpha)| \left|\int_{\xi}^{c_2} g(x, \alpha) \, dx\right|$$
Using the uniform boundedness of the integral of $g$ and the uniform convergence of $f(x,\alpha)$ to zero, we can show that for all $c_2 > c_1$ sufficiently large (and independent of $\alpha \in (0,A]$), the Cauchy criterion for uniform convergence is satisfied.
A: You can use Dirichlet's Criterion:
Let $f\colon (a, b) \to \mathbb{R}$ a monotone function on $(a, b)$ and $g\colon (a, b)\to \mathbb{C}$ an integrable function in $(a, t]$, for each $t\in (a, b)$. If $\displaystyle \lim_{x\to b} f(x) = 0$, and $\int_{a}^{t} g(x) dx$ is bounded for each $t \in (a, b)$, then $\int_{a}^{\to b} fg $ converges.
Taking $a = 0$, $b = \infty$, $f = \frac{1}{\alpha^2 + x^2}$ and $g = \sin \alpha x$ we have $$\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha^2 + x^2} \quad \text{converges}$$
