# Question c8269

##### 1 Answer
Jun 8, 2015

${z}^{\alpha} / \left\{{z}^{n} + 1\right\}$ has simple poles at

z_k=exp(ipi(1/n+{2k}/n), \quad \quad k=0, 1, 2, ...\quad (n-1)#

Note that $\frac{d}{\mathrm{dz}} \left[{z}^{n} + 1\right] = n {z}^{n - 1}$ is analytic for $n > 1$ and is not zero at ${z}_{k}$. Therefore, ${z}_{k}$ is a zero of order 1 of $\left({z}^{n} + 1\right)$ which lets us write:

$\left({z}^{n} + 1\right) = \left(z - {z}_{k}\right) g \left(z\right)$

where $g \left(z\right)$ is analytic and $g \left({z}_{k}\right) \setminus \ne 0$. It follows that

${z}^{\alpha} / \left\{{z}^{n} + 1\right\} = \frac{{z}^{\alpha} / g \left(z\right)}{\left(z - {z}_{k}\right)}$

The residue at ${z}_{k}$ is

${B}_{k} = {\left({z}_{k}\right)}^{\alpha} / g \left({z}_{k}\right)$

To find $g \left({z}_{k}\right)$ use,

$\frac{d}{\mathrm{dz}} \left[{z}^{n} + 1\right] = \frac{d}{\mathrm{dz}} \left[\left(z - {z}_{k}\right) g \left(z\right)\right]$

$n {z}^{n - 1} = g \left(z\right) + \left(z - {z}_{k}\right) g ' \left(z\right)$

Sub $z = {z}_{k} \setminus \quad \setminus \implies \setminus \quad g \left({z}_{k}\right) = n {\left({z}_{k}\right)}^{n - 1}$

$\setminus \implies \setminus \quad {B}_{k} = {\left({z}_{k}\right)}^{\alpha} / \left\{n {\left({z}_{k}\right)}^{n - 1}\right\} = \frac{1}{n} {\left({z}_{k}\right)}^{1 + \alpha - n}$

Now we must choose an integration path. The problem is choosing a path that won't have a singularity on it for any value of $n$. I am going to use the path described below.

${C}_{1} = {\int}_{0}^{\infty} {x}^{\alpha} / \left({x}^{n} + 1\right) \mathrm{dx}$

${C}_{2} = {\int}_{\infty}^{0} {\left(R {e}^{i \frac{2 \pi}{n}}\right)}^{\alpha} / \left({\left(R {e}^{i \frac{2 \pi}{n}}\right)}^{n} + 1\right) \mathrm{dR} = {e}^{i 2 \pi \frac{\alpha}{n}} {\int}_{\infty}^{0} {R}^{\alpha} / \left({R}^{n} + 1\right) \mathrm{dR}$

Note that
${C}_{2} = - {e}^{i 2 \pi \frac{\alpha}{n}} {C}_{1}$

Before evaluating ${C}_{R}$ consider

$| {z}^{\alpha} / \left({z}^{n} + 1\right) | \le q {R}^{\alpha} / \left\{{R}^{n} - 1\right\}$

So $| {C}_{R} | \setminus \le q {R}^{\alpha} / \left\{{R}^{n} - 1\right\} 2 \pi R = 2 \pi \frac{1}{\left\{{\left(R\right)}^{n - \left(\alpha + 1\right)}\right\} - {R}^{- \alpha - 1}}$

We are given that $n > \left(\alpha + 1\right)$, Therefore ${\lim}_{R \setminus \rightarrow \infty} {C}_{R} = 0$

Note that I have chosen $\theta$ in the diagram so that only one singularity is enclosed. Therefore we only have to consider the first residue, ${B}_{0}$.

${C}_{1} + {C}_{2} + {C}_{R} = 2 \pi i {B}_{0}$

Substitute values and solve for ${C}_{1}$

${C}_{1} \left(1 - {e}^{i 2 \pi \frac{\alpha}{n}}\right) + 0 = 2 \pi i \frac{1}{n} {\left({z}_{0}\right)}^{1 + \alpha - n}$

Sub ${z}_{0}$ To get the answer:

${\int}_{0}^{\infty} {x}^{\alpha} / \left({x}^{n} + 1\right) \mathrm{dx} = {C}_{1} = \frac{2 \pi i}{n} \frac{1}{1 - {e}^{i 2 \pi \frac{\alpha}{n}}} {e}^{i \pi \frac{1 + \alpha - n}{n}}$