Complex Integration - $\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx$

Question

Exercise :

Show that :

$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \frac{4\pi}{5}\sin\bigg(\frac{2\pi}{5}\bigg)$$

Attempt :

$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^3 + x^2 + x + 1}dx $$

$$x^4 + x^3 + x^2 + x + 1=0 \Leftrightarrow x = \{-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5} \}$$

So, the function $f(z) = \frac{1}{x^4 + x^3 + x^2 + x + 1}$ has poles at the points :

$$\{-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5} \}$$

Now, I know you have to integrate through a closed curve $C$ and on a line $γ_R$ and then continue on with residues for the poles that reside in this curve, but I am stuck on how to apply it here and I also miss it a bit on how to split the integral for the curve and the line. Most examples I've saw get simpler due to the even function trick, but that cannot be applied here.

I would really appreciate a thorough solution and explanation, since I've just started working on generalized integrals and I have to clear my mind on them.

Thanks for your time !

Answer 1

Using the upper part of the complex plane we get:

$\displaystyle \int_{-\infty}^{+\infty}\frac{x-1}{x^5-1}dx=i2\pi Res(\frac{x-1}{x^5-1},e^{i2\pi/5})+ i2\pi Res(\frac{x-1}{x^5-1},e^{i4\pi/5})$

$\displaystyle = i2\pi\frac{x-1}{x^5-1}(x-e^{i2\pi/5})|_{x\to e^{i2\pi/5}}+ i2\pi\frac{x-1}{x^5-1}(x-e^{i4\pi/5})|_{x\to e^{i4\pi/5}} $

$\displaystyle =\frac{\pi}{4}\frac{1}{\sin(\frac{\pi}{5})\sin(\frac{3\pi}{5})\sin(\frac{4\pi}{5})}=\frac{4\pi}{5}\sin\frac{2\pi}{5}$

Answer 2

Brute force also helps: $$\int\limits_{-\infty}^{+\infty}\frac{x-1}{x^5-1}dx=\int\limits_{-\infty}^{+\infty}\frac{1}{x^4+x^3+x^2+x+1}dx=$$ $$=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(x^2+\frac{1}{x^2}+x+\frac{1}{x}+1\right)}dx=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(\left(x+\frac{1}{x}\right)^2+x+\frac{1}{x}-1\right)}dx=$$ $$=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(x+\frac{1}{x}-\frac{-1+\sqrt5}{2}\right)\left(x+\frac{1}{x}-\frac{-1-\sqrt5}{2}\right)}dx=$$ $$=\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x^2-\frac{\sqrt5-1}{2}x+1\right)\left(x^2+\frac{\sqrt5+1}{2}x+1\right)}dx=$$ $$=\frac{1}{\sqrt5}\int\limits_{-\infty}^{+\infty}\left(\frac{x+\frac{\sqrt5+1}{2}}{x^2+\frac{\sqrt5+1}{2}x+1}-\frac{x-\frac{\sqrt5-1}{2}}{x^2-\frac{\sqrt5-1}{2}x+1}\right)dx=$$ $$=\frac{1}{\sqrt5}\int\limits_{-\infty}^{+\infty}\left(\frac{\frac{\sqrt5+1}{4}}{x^2+\frac{\sqrt5+1}{2}x+1}-\frac{\frac{\sqrt5-1}{4}}{x^2-\frac{\sqrt5-1}{2}x+1}\right)dx=$$ $$=\frac{\sqrt5+1}{4\sqrt5}\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x+\frac{\sqrt5+1}{4}\right)^2+1-\frac{6+2\sqrt5}{16}}dx-\frac{\sqrt5-1}{4\sqrt5}\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x-\frac{\sqrt5-1}{4}\right)^2+1-\frac{6-2\sqrt5}{16}}dx=$$ $$=\pi\left(\frac{\sqrt5+1}{4\sqrt5\sqrt{\frac{10-2\sqrt5}{16}}}-\frac{\sqrt5-1}{4\sqrt5\sqrt{\frac{10+2\sqrt5}{16}}}\right)=\frac{\pi}{\sqrt5}\left(\frac{\cos36^{\circ}}{\sin36^{\circ}}-\frac{\sin18^{\circ}}{\cos18^{\circ}}\right)=$$ $$=\frac{\pi\cos54^{\circ}}{\sqrt5\sin36^{\circ}\cos18^{\circ}}=\frac{\pi}{\sqrt5\cos18^{\circ}}.$$