Here is another solution:
Consider the integral
$I(\alpha) = \int_{0}^{\infty} \frac{\sin (\alpha x)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{\alpha \sin x}{\alpha^2+x^2} \, dx.$
Differentiating $I(\alpha)$ with the first equality, we have
\begin{align*} I'(\alpha) &= \int_{0}^{\infty} \frac{x \cos (\alpha x)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{x \cos x}{\alpha^2+x^2} \, dx. \end{align*}
Differentiating once again, we have
\begin{align*} I''(\alpha) &= -\int_{0}^{\infty} \frac{2\alpha x \cos x}{(\alpha^2+x^2)^2} \, dx = \left[ \frac{\alpha \cos x}{\alpha^2+x^2} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{\alpha \sin x}{\alpha^2+x^2} \, dx \\ &= -\frac{1}{\alpha} + I(\alpha). \end{align*}
Thus $I$ satisfies the differential equation
$ I'' - I = -\frac{1}{\alpha}. \tag{1}$
To solve this equation, we let
$ I(\alpha) = u e^{\alpha}. $
Plugging this to $(1)$ and multiplying $e^{\alpha}$ to both sides, we obtain
$ (u'e^{2\alpha})' = -\frac{1}{\alpha}e^{\alpha}. $
Thus integrating both sides, we have
$ u'e^{2\alpha} = -\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}, $
where
$\mathrm{Ei}(\alpha) = PV \int_{-\infty}^{\alpha} \frac{e^{t}}{t} \, dt$
is the exponential integral function. Then
$ u' = -e^{-2\alpha}\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}e^{-2\alpha} $
and hence
\begin{align*} u &= \int \left( -e^{-2\alpha}\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}e^{-2\alpha} \right) \, d\alpha \\ &= \frac{1}{2}e^{-2\alpha} \mathrm{Ei}(\alpha) - \int \frac{e^{-\alpha}}{2\alpha} \, d\alpha + c_{1}e^{-2\alpha} + c_{2} \\ &= \frac{1}{2}e^{-2\alpha} \mathrm{Ei}(\alpha) - \frac{1}{2}\mathrm{Ei}(-\alpha) + c_{1}e^{-2\alpha} + c_{2}. \end{align*}
Therefore it follows that
$ I(\alpha) = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2} + c_{1}e^{-\alpha} + c_{2} e^{\alpha} $
for some $c_1$ and $c_2$. To determine $c_1$ and $c_2$, observe that
$\mathrm{Ei}(\alpha) \sim c + \log |\alpha|$
near $\alpha = 0$. (In fact, we have $c = \gamma$.) Thus taking $\alpha \to 0$,
$ 0 = I(0) = c_1 + c_2. $
This shows that we may write
$ I(\alpha) = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2} + c \sinh \alpha. $
But L'hospital's rule shows that
$ \mathrm{Ei}(\alpha) \sim \frac{e^{\alpha}}{\alpha} $
as $|\alpha| \to \infty$. Thus $ I(\alpha) \sim c \sinh \alpha$ as $\alpha \to \infty$. But it is clear that $I(\alpha)$ is bounded:
$ \left|I(\alpha)\right| \leq \int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2}. $
Therefore $c = 0$ and we have
$ \int_{0}^{\infty} \frac{\sin (\alpha x)}{1+x^2} \, dx = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2}. $