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[SOLVED] Contour integration

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\int_{-\infty}^{\infty}\frac{x - 1}{x^5 - 1}dx.
$$

Taking a look at the expression $f(z) = \dfrac{z - 1}{z^5 - 1}$ we see that the poles of $f$ are the zeros of $g(z) = z^5 - 1 = 0$.
Let $z = re^{i\theta}$ where $\theta\in (-\pi, \pi)$.
Then
$$
r^5e^{5i\theta} = 1
$$
and $e^{2i\pi k} = 1$ where $k\in\mathbb{Z}$.
So
$$
r^5e^{5i\theta} = e^{2i\pi k},
$$
Thus, we have that $r^5 = 1$ so $r = 1$ and $5\theta = 2\pi k$ so $\theta = \dfrac{2\pi k}{5}$.
Then
$$
z_j = \exp{\left(\pm\dfrac{2\pi i}{5}\right)}, \exp{\left(\pm\dfrac{4\pi i}{5}\right)}, 1.
$$
The only zeros in the upper half plane are $z = \exp{\left(\dfrac{2\pi i}{5}\right)}, \exp{\left(\dfrac{4\pi i}{5}\right)}$.
$g'(z) = 5z^4$ which is zero iff. $z = 0$.
Thus $1/g$ has only simple poles at $z_j$.
\begin{align}
\int_{-\infty}^{\infty}\frac{x - 1}{x^5 - 1}dx &= 2i\pi\sum_{z_j \ \text{upper half plane}}\text{Res}_{z = z_j}g(z)\notag\\
&= \frac{2\pi i}{5}\left(\exp{\left(-\dfrac{8\pi i}{5}\right)} + \exp{\left(-\dfrac{6\pi i}{5}\right)}\right)\notag\\
\end{align}

So I get to here but I can't simplify it down to Mathematica's answer. Plus, when I get the numerical solution of my answer and Mathematica's, they aren't the same.
What is wrong?

---------- Post added at 15:28 ---------- Previous post was at 15:00 ----------

Does f(z) need to be defined by removing the removable singularity?

$$
f(z) = \dfrac{1}{z^4 + z^3 + z^2 + z + 1}
$$
and then
$$
g(z) = z^4 + z^3 + z^2 + z + 1 = 0
$$
If this is the case, how would (see below) be solved?
$$
r^4e^{4i\theta} + r^3e^{3i\theta} + r^2e^{2i\theta} + re^{i\theta} = e^{i\pi + 2i\pi k}
$$
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
It looks as though you are using the wrong formula for the residues. For a function of the form $\dfrac{h(z)}{g(z)}$, the residue at a pole $z_0$ is $\dfrac{h(z_0)}{g'(z_0)}.$ In this case, the function is $\dfrac{z-1}{z^5-1}$, so the residue is $\dfrac{z_0-1}{5z_0^4}.$

You will find the calculation much easier if you write $\omega = e^{2\pi i/5}$ (and remember all the time that $\omega^5=1$). Then the poles are at $\omega$ and $\omega^2$. The residues are $\dfrac{\omega-1}{5\omega^4}$ and $\dfrac{\omega^2-1}{5\omega^3}.$ Their sum simplifies to $\tfrac15(-\omega+\omega^4)$. But $\omega^4$ is the complex conjugate of $\omega$, so this simplifies further to $-\tfrac25i\sin\tfrac{2\pi}5.$ That gives the value of the integral as $\tfrac{4\pi}5\sin\tfrac{2\pi}5.$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
It looks as though you are using the wrong formula for the residues. For a function of the form $\dfrac{h(z)}{g(z)}$, the residue at a pole $z_0$ is $\dfrac{h(z_0)}{g'(z_0)}.$ In this case, the function is $\dfrac{z-1}{z^5-1}$, so the residue is $\dfrac{z_0-1}{5z_0^4}.$

You will find the calculation much easier if you write $\omega = e^{2\pi i/5}$ (and remember all the time that $\omega^5=1$). Then the poles are at $\omega$ and $\omega^2$. The residues are $\dfrac{\omega-1}{5\omega^4}$ and $\dfrac{\omega^2-1}{5\omega^3}.$ Their sum simplifies to $\tfrac15(-\omega+\omega^4)$. But $\omega^4$ is the complex conjugate of $\omega$, so this simplifies further to $-\tfrac25i\sin\tfrac{2\pi}5.$ That gives the value of the integral as $\tfrac{4\pi}5\sin\tfrac{2\pi}5.$
How do the sums equal $\frac{1}{5}(-\omega+\omega^4)$?

I keep getting $\frac{1}{5}\left(\dfrac{\omega^3 - 1}{\omega^4}\right)$

Never mind.