Compute the integral using the residue theorem: $\int_{-\infty}^\infty \frac{x^2}{x^6 + 1}dx$.
If we let $\gamma_R$ be the line from $-R$ to $R$, and $\gamma_C$ be the upper half circle, and integrate ccw, we have
$$\int_{\gamma_C \cup \gamma_R} \frac{x^2}{x^6+1}dx = \int_{-R}^R \frac{x^2}{x^6+1}dx + \int_{\gamma_C}\frac{x^2}{x^6 + 1}dx.$$
Now, I need to calculate the residues of the LHS. But it seems really difficult! I see that the poles in our contour are at $x_1 = e^{\pi/6}$, $x_2 = e^{3\pi/6}$, $x_3 = e^{5\pi/6}$. Now, here is where I am getting lost:
We have $\text{res}_{x_1} = \lim_{x\to e^{\pi/6}} (x-e^{\pi/6})\frac{x^2}{x^6 + 1}.$ But if I try to split up the $x^6 + 1$ in the bottom of the fraction here, I get a bunch of terms $(x-e^{3\pi/6})(e-e^{5\pi/6})…$ etc., and that seems very messy to calculate for each pole! Is there an easier/cleaner way to do this?
Best Answer
If you have $f(z)/g(z)$ with a simple pole in $a\in\mathbb{C}$, the residue in $a$ is simply $$\frac{f(a)}{g'(a)}.$$
Proof: since $g(a)=0$ the residue is $$\lim_{z\to a} \frac{f(z)}{g(z)}(z-a)=\lim_{z\to a}f(z)\frac{z-a}{g(z)-g(a)}$$
So in your case the residue is $\frac{a^2}{6a^5}=\frac{1}{6}a^{-4}$ and now substitute the correct values of $a$.