I'm supposed to evaluate:
$$ \int_0^{\infty} \frac{\sin x}{x(k^2x^2 +1)} dx $$
Attempt
Consider $ \int_0^{\infty} \frac{e^{iz}}{x(k^2x^2 +1)} dz $
Simple poles at $z = \pm \frac{i}{k} $, simple pole at z = 0.
Consider contour the following contour:
Residue at $z=\frac{i}{k}$ is $-\frac{1}{2} e^{-\frac{1}{k}} $
Residue at $z=0$ is $0$.
By Jordan's lemma, the integral along the chord disappears as the circle blows up. Thus only integral along real axis remains.
$$ \int_{-\infty}^{\infty} \frac{e^{ix}}{x(k^2x^2+1)} = 2\pi i \times (-\frac{1}{2} e^{-\frac{1}{k}}) $$
Looks good, as due to parity the cosine part disappears.
Thus, $ \int_0^{\infty} \frac{\sin x}{x(k^2x^2 +1)} dx = -\frac{\pi}{2}e^{-\frac{1}{k}} $