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$$

\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}

$$

\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: