Integrating rational functions of sine using residues Can anyone point me to a book that discusses integrals of the following type?
$$\int_{\mathbb{R}}^{}\frac{\cos(ax)dx}{1+x^2}$$
 A: Actually, Jordan's lemma is not needed here, at least if assuming $a$ is real. Since the degree of the denominator is two, we can just write $\cos ax = \operatorname{Re} e^{iax}$, and if $a > 0$ and $z$ is in the upper halfplane, $$|e^{iaz}| = |e^{ia(x+iy)}| = e^{-ay} < 1.$$
Hence, choosing $f(z) = \dfrac{e^{iaz}}{1+z^2}$ and a contour $\gamma = [-R,R] \cup C_R^+$, where $C_R^+$ is a semi-circle in the upper halfplane of radius $R$. By the residue theorem it follows that
$$ \int_{-R}^R \dfrac{e^{iaz}}{1+z^2}\,dz + \int_{C_R^+} \dfrac{e^{iaz}}{1+z^2}\,dz = 2\pi i \operatorname{Res}_{z=i}(f(z)).$$
By the estimate above,
$$\left|\dfrac{e^{iaz}}{1+z^2}\right| \le \frac{1}{R^2-1}$$
on $C_R^+$, so the integral over $C_R^+$ tends to $0$ as $R\to\infty$ by the "ML-inequality". Summing up, you will see that
$$\int_{-\infty}^\infty \dfrac{\cos ax}{1+x^2}\,dx = \operatorname{Re}(2\pi i \operatorname{Res}_{z=i}(f(z))).$$
If $a < 0$, you can do the same with a semicircle in the lower halfplane (or just use the fact that $\cos(ax) = \cos(-ax)$).
Similar estimates work for integrals of the type
$$\int_{-\infty}^\infty \cos ax \frac{P(x)}{Q(x)}\,dx
\quad\text{or}\quad
\int_{-\infty}^\infty \sin ax \frac{P(x)}{Q(x)}\,dx
$$
where $P$ and $Q$ are polynomials and $\deg Q \ge \deg P + 2$. If $\deg Q = \deg P + 1$ you do need some variant of Jordan's lemma, this should be covered in any textbook on complex analysis.
