A nasty integral

In this post, we discuss how certain integrals of rational functions can be computed.

Example 1. $$\int_{0}^{\infty} \frac{x^{8} - 4x^{6} + 9x^{4} - 5x^{2} + 1}{x^{12} - 10x^{10} + 37x^{8} - 42x^{6} + 26x^{4} - 8x^{2} + 1} \, \mathrm{d}x = \frac{\pi}{2}.$$ This is Problem 11148 of American Mathematical Monthly, Vol.112, April 2005.

Proof. For each $f(z) \in \mathbb{C}[z]$, we define $f^*(z) = \overline{f(\bar{z})}$. This is the same as taking conjugate to the coefficients appearing in $f(z)$. Let $$\begin{align*} P(z) &= z^8-4 z^6+9 z^4-5 z^2+1, \\ Q(z) &= z^{12}-10 z^{10}+37 z^8-42 z^6+26 z^4-8 z^2+1. \end{align*}$$ The key observation is that all the zeros of the polynomial $$q(z) = z^3-(2+i) z^2-(1-i) z+1$$ lies in the upper half-plane $\mathbb{H} = \{ z \in \mathbb{C} : \operatorname{Im}(z) \gt 0\}$, and $$Q(z) = q(z)q^*(z)q(-z)q^*(-z)$$ is a factorization of $Q(z)$ into coprime factors. Now let $p(z)$ be the unique polynomial such that $\deg p \lt \deg q$ and $$P(z) \equiv p(z) q^*(z)q(-z)q^*(-z) \pmod{q(z)}.$$ Such $p(z)$ exists because $q(z)$ and $q^*(z)q(-z)q^*(-z)$ are coprime. Also, such $p(z)$ can be computed using the extended Euclidean algorithm. After a tedious computation, it follows that $$p(z) = \frac{-i z^2-(1-2 i) z+1}{4}.$$ Then by the symmetry and the uniqueness of partial fraction decomposition, it follows that $$\frac{P(z)}{Q(z)} = \frac{p(z)}{q(z)} + \frac{p^*(z)}{q^*(z)} + \frac{p(-z)}{q(-z)} + \frac{p^*(-z)}{q^*(-z)}.$$ Moreover, using the fact that all the zeros of $q(z)$ lies in $\mathbb{H}$, we can invoke Cauchy's integral theorem to write $$\begin{align*} \operatorname{PV}\!\! \int_{-\infty}^{\infty} \frac{x^n}{q(x)} \, \mathrm{d}x &= \lim_{R\to\infty} \int_{-R}^{R} \frac{x^n}{q(x)} \, \mathrm{d}x \\ &= \lim_{R\to\infty} \int_{-\pi}^{0} \frac{i (Re^{i\theta})^{n+1}}{q(Re^{i\theta})} \, \mathrm{d}\theta \\ &= \begin{cases} 0, & n \lt \deg q - 1, \\ \pi i, & n = \deg q - 1. \end{cases} \end{align*}$$ Therefore we conclude that $$\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} \, \mathrm{d}x = 4 \operatorname{Re}\left[ \operatorname{PV}\!\! \int_{-\infty}^{\infty} \frac{p(x)}{q(x)} \, \mathrm{d}x \right] = 4 \pi i \left( [z^2] p(z) \right) = \pi.$$ The answer is half of the above integral, hence the claim follows. $\square$

Here is another example.

Example 2. $$\int_{0}^{\infty} \frac{x^{14} - 15x^{12} + 82x^{10} - 190x^{8} + 184x^{6} - 60x^{4} + 16x^{2}}{x^{16} - 20x^{14} + 156x^{12} - 616x^{10} + 1388x^{8} - 1792x^{6} + 1152x^{4} - 224x^{2} + 16} \, \mathrm{d}x = \frac{\pi}{2}.$$

Proof. Apply the same argument as above with the choices $$q(z) = z^4+(3-i) z^3-(1+3 i) z^2-(6+2 i) z-2$$ and $$p(z) = \frac{-i z^3-3 i z^2-2 i z}{4}.$$ We assure the reader that all the zeros of $q(z)$ lie in the upper half-plane, and the denominator of the integral admits the same form of factorization into coprime factors as before. $\square$