complex analysis - How to evaluate $int_{-infty}^{infty} frac{1}{x^6+1},dx$?

I am trying to understand how to evaluate the following integral.

$$\int_{-\infty}^{\infty} \dfrac{1}{x^6+1}dx$$

We start by considering $f(z) = \dfrac{1}{z^6+1}dz.$ Then we find that $z_k = \exp \left (i \left (\frac{\pi}{6}+\frac{\pi k}{3}\right)\right )$, and simple poles occur when $k=0,1,2,3,4,5$.

Then

$$\text{Res}(f,z_k)= \dots =\dfrac{1}{6}\lim_{z \to z_k} z^{-5}= \dfrac{1}{6} \exp \left( -\frac{5 \pi i}{6} (1 +2k) \right)$$

I think that in the missing part, the following formula is used,

$$\text{Res}(f,c)=\frac{1}{(n-1)!}\lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}} ((z-c)^n f(z)) $$

that $n=4$ and that a cancellation occurs (otherwise it would be really messy), but I'm not sure how to do it. How do I proceed?

Thanks for your help.

Answer

This is rather peculiar: the denominator factors to

$$ \prod_{k=0}^5 (z-z_k), $$
and the function has no real roots, so you take the three in the upper half-plane and find the residues at their poles. The three roots are simple, so the residue formula in its most basic form should be used, and you find something like
$$ 2\pi i \left( \prod_{k=1}^5 \frac{1}{z_0-z_k} + \prod_{\substack{k=0 \\ k \neq 1}}^5 \frac{1}{z_1-z_k} + \prod_{\substack{k=0 \\ k \neq 2}}^5 \frac{1}{z_2-z_k} \right). $$
But as far as I know, there is no simpler way of doing this: you can make various simplifications, but you still end up having to push quite a lot of algebra through.

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However, you don't need to do it that way: first, the integrand is symmetric, so the integral is twice the integral over $[0,\infty)$. Now substitute $x=y^{1/6}$, so
$$ I = 2 \int_0^{\infty} \frac{1}{6} \frac{y^{-5/6}}{1+y} \, dy $$
There are a variety of ways of doing this integral, probably the easiest being to substitute $u=y/(1+y)$. Then $y= u/(1-u)= 1+\frac{1}{u-1}$, $dy = \frac{1}{(1-u)^2} \, du $, so

$$ I = \frac{1}{3} \int_0^1 \left( \frac{u}{1-u} \right)^{-5/6} \frac{1-u}{1-u+u} \frac{1}{(1-u)^2} \, du = \frac{1}{3} \int_0^1 u^{-5/6} (1-u)^{-1/6} \, du = \frac{1}{3}\frac{\Gamma(1/6)\Gamma(5/6)}{\Gamma(1)} = \tfrac{1}{3}\pi\csc{\tfrac{1}{6}\pi}, $$
using the Beta function.