Integrating with Infinite boundaries

[SALAAH_BEDDIAF]

Homework Statement

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

The Attempt at a Solution

This is actually a piece of work from a complex analysis module (not sure if it belongs in this part of the forum or in the analysis section)

I know that for any infinite bounded integrals you separate them to be $$\lim_{c\to -\infty} \int_c^0f(x)dx + \lim_{c\to +\infty}\int_0^cf(x)dx$$
and I have factorised $x^5-1 = (x-1)(x^4 + x^3 + x^2 + x + 1)$

Will I be integrating this as normal or is there a different way to do this if so please help me start :(.

 

[vela]

Since you found this problem in the context of complex analysis, you're probably meant to evaluate the integral by associating it with a complex integral, which you can evaluate using the residue theorem.

 

[Mark44]

SALAAH_BEDDIAF, this is the correct forum section. I deleted your post in the other section.

 

[SALAAH_BEDDIAF]

So I understand I will be evaluating this integral using Cauchy's Residue Theorem so I must find the singularities in the function, this is when $(x-1)(x^4 + x^3 + x^2 + x + 1) = 0$, from $(x-1)$ we get a singularity to be 1, but I'm having difficulty finding $(x^4 + x^3 + x^2 + x + 1)$ to be 0

 

[Mark44]

The solutions of x⁵ - 1 = 0 are the 5th roots of unity. One of the roots is 1, which you already know. The other four are complex.

 

[SALAAH_BEDDIAF]

The solutions of x⁵ - 1 = 0 are the 5th roots of unity. One of the roots is 1, which you already know. The other four are complex.

so the roots of $(x^4 + x^3 + x^2 + x + 1)$ are also the remaining roots of $(x^5-1)$?

 

[vela]

Yes.

 

[SALAAH_BEDDIAF]

Okay, so I have all the roots now, as these roots are very difficult to be represented by fractions, can I use the approximate roots to find the residues or will I have to use the exact value?

 

[Mark44]

If you write them in polar form, they're very simple.

 

[HallsofIvy]

All n roots of $z^n- 1= 0$ lie on the circle |z|= 1 and are equally spaced around it. Since the denominator is fifth degree (odd) it has roots 1 and -1 but no roots on the imaginary axis. Of course, to use residues you need to integrate around a closed path. I would recommend integrating from "-iR" to "iR", for R> 1, along the imaginary axis, then on a semicircle back to "-iR". Finally, let R go to infinity.

 

[Mark44]

All n roots of $z^n- 1= 0$ lie on the circle |z|= 1 and are equally spaced around it. Since the denominator is fifth degree (odd) it has roots 1 and -1 ...

-1 isn't one of the roots. If n were even, it would be, but n is odd in this equation.