Computing integrals for trigonometric-polynomial quotients

Description

Here, we describe the integration procedure for integrals of the form:

$\mathrm{P}\mathrm{V}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{P(x)\sin x}{Q(x)}$

The $\sin$ may be replaced by $\cos$.

Here, we assume that the polynomial $Q$ does not have any zeros on the real axis (except possibly at the origin).

Solution by semicircular contour integration

Further information: semicircular contour integration method

This is typical, for instance, when the integrand is of the form:

$f(x):=\frac{\sin x}{Q(x)}$

where the degree of $Q$ is strictly greater than one, and $Q$ has no real zeros. We use the complex-valued function $e^{iz}/Q(z)$. In this case, Jordan's lemma (or the weaker version, the so-called [semicircular contour theorem]]) tells us that the integral along large semicircles tends to zero, so:

$\mathrm{P}\mathrm{V}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{Q(x)}=\mathrm{I}\mathrm{m}((2\pi i)\sum \mathrm{r}\mathrm{e}\mathrm{s}(f;z_j))$

where the sum is taken over all the zeros $z_j$ of $f$ that lie in the upper half-plane.

Note that we can drop the $\mathrm{P}\mathrm{V}$ (Cauchy principal value symbol), since the integral is absolutely convergent in this case. Also,if $Q$ has only simple poles, so that we have:

$Q(z):=\prod _{j=1}^n(z-z_j)$

Then, we get the formula:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{Q(x)}=2\pi \mathrm{R}\mathrm{e}(\sum _{j=1}^n\frac{\sin (z_j)}{{\prod }_{k\ne j}(z_j-z_k)})$