# Computing an integral where the poles of the integrand are the roots of unity

I am trying to compute the following integral: $$\int_0^{2\pi} \frac{y}{y^n-1} dy$$

I've tried to decompose $y^n-1$ into $(y-1)(y-e^{i\theta})(y-e^{i2\theta})...(y-e^{i(n-1)\theta})$ but I don't know what to do with this factorization. I've read some others similar questions with the answers but I don't know if the same methods apply.

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You are integrating on an interval and I am not sure what the roots of unity have do with anything. Also, the integral doesn't exist. – SteveH Jun 30 '11 at 22:24
That is weird. I was originally trying to integrate $\int_{S^1} dz/(z^n-a)$ and that's how I got this integral (I didn't include the factor $a^{2/(n-1)}$ that I got from the substitution I did for simplicity). $S^1$ is the unit circle. – user786 Jun 30 '11 at 22:36
OK. Assume $|a|\not=1$. Treat the cases $|a|>1$ and $|a|<1$ separately. One case is easy. Try the other case by thinking Cauchy's Integral formula, small circles around the poles and a suitable contour that incorporates these small circles and the unit circle. – SteveH Jun 30 '11 at 22:44
Thank you! I'll come back if I still can't make substantial progress. – user786 Jun 30 '11 at 22:47
The integral in your question isn't the integral you want. You want to replace $y$ with $z = e^{iy}$ in the integrand or something similar. – Qiaochu Yuan Jun 30 '11 at 23:23

## 2 Answers

Hint: Let $\xi_k, k=1,2,\ldots,n,$ be any one of the zeros of $z^n-a$, and $f(z)=1/(z^n-a)$, so $\xi_k$ is a simple pole of $f(z)$ (assuming that $a\neq0$). Then by l'Hospital $$Res(f(z),\xi_k)=\lim_{z\to\xi_k}\frac{z-\xi_k}{z^n-a}= \lim_{z\to\xi_k}\frac{1}{nz^{n-1}}=\frac{1}{n\xi_k^{n-1}}=\frac{\xi_k}{n\xi_k^n}=\frac{\xi_k}{na}.$$

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Thanks for the hint. That's what I wanted to do but which poles are located in the upper half plane? – user786 Jul 1 '11 at 20:09
Why worry about the upper half plane? Aren't you integrating along the unit circle? Read the comment by SteveH. – Jyrki Lahtonen Jul 1 '11 at 20:53

Let $f(z) = \frac{z}{z^n - 1}$. You're looking for $\oint_{\gamma} f(z) dz$ taken over a contour that is outside of the unit circle, loosely speaking. We will use the Residue Theorem to compute the integral. Define $\omega_k = e^{2 \pi i k / n}$ for $k = 0, \dots, n-1$ to be the roots of unity, which are also the poles of $f(z)$. We calculate the residues using the factorization $z^n - 1 = \prod_{k=0}^{n-1} (z-\omega_k)$ as follows:

$$R_k = \lim_{z\to \omega_k} (z-\omega_k)f(z) = \omega_k \prod_{j \ne k} (\omega_k - \omega_j)^{-1} = \omega_k \prod_{l = 1}^{n-1} \left[ \omega_k^{-1} ( 1 - \omega_l)^{-1} \right] = \omega_k^2 R_0 .$$

Note that the above uses $\omega_k^{-n} = 1$ and the index substitution $l = j-k$. We could use the geometric formula to calculate $R_0 = 1/(n-1)$, but it's actually unnecessary because we know that $\sum \omega_k^2 = 0$ (for $n > 2$), since it is the $z^2$ coefficient of the polynomial $z^n-1$. Hence the integral vanishes.

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Thanks for your answer. It may be trivial but why is the sum of the squares of the nth roots equal to 0? – user786 Jul 1 '11 at 20:10
To that end: recall the connection between the zeros of a polynomial and its coefficients. If $$p(z)=(z-z_1)(z-z_2)\cdots(z-z_n)=z^n+a_1z^{n-1}+a_2z^{n-2}+\cdots+a_n,$$ then the sum of the zeros is the negative of $a_1$ et cetera. The sum of squares is a combination of the two highest coefficients. Not sure that you need to look at that, though? – Jyrki Lahtonen Jul 1 '11 at 21:56