Contour Integration I'm having trouble with the step highlighted in green. As the angle $2\pi /n$is maintained as R goes to infinity, I don't see how you end up just integrating on the x axis.


 A: Since the notes you attached are a little cavalier, I have added the details for the problem. Let $f(z) = \frac{1}{1+z^n}$ and $\Gamma$ is the arc. From the contour, we have
$$
\int_0^{\infty}\frac{dx}{1 + x^n}  = \lim_{R\to\infty}\biggl[\int_0^R+\int_{\Gamma}+\int_R^0\biggr]f(z)dz = 2i\pi\sum_j\operatorname{Res}\{f(z);z_j\}
$$
The only pole in the sector is then $z = e^{i\pi/n}$. The residue is then
$$
\lim_{z\to e^{i\pi/n}}\frac{z-e^{i\pi/n}}{1+z^n}=\lim_{z\to e^{i\pi/n}}\frac{1}{nz^{n-1}}=\frac{-e^{i\pi/n}}{n}
$$
Therefore, we now have
$$
\frac{-2i\pi e^{i\pi/n}}{n} = \int_0^{\infty}\frac{dz}{1+z^n}+\lim_{R\to\infty}\int_0^{2\pi/n}\frac{iRe^{i\theta}}{1+R^ne^{in\theta}}d\theta+\int_{\infty}^0\frac{dz}{1+z^n}
$$
where, for the third integral, $z = re^{2i\pi/n}$, $r\in[0,R]$, and $dz = e^{2i\pi/n}dr$. Then the third integral can be written as
$$
-\int_0^{\infty}\frac{e^{2i\pi/n}}{1+r^n}dr
$$
Since $r$ is a dummy variable, let $r = z$.
\begin{align}
\int_0^{\infty}\frac{dz}{1+z^n} &= -\lim_{R\to\infty}\int_0^{2\pi/n}\frac{iRe^{i\theta}}{1+R^ne^{in\theta}}d\theta+\int_0^{\infty}\frac{e^{2i\pi/n}}{1+z^n}dz - \frac{2i\pi e^{i\pi/n}}{n}\\
(1-e^{2i\pi/n})\int_0^{\infty}\frac{dz}{1+z^n}&= -\lim_{R\to\infty}\int_0^{2\pi/n}\frac{iRe^{i\theta}}{1+R^ne^{in\theta}}d\theta - 
\frac{2i\pi e^{i\pi/n}}{n}
\end{align}
Let's consider the first integral only now.
$$
\Biggl\lvert\int_0^{2\pi/n}\frac{iRe^{i\theta}}{1+R^ne^{in\theta}}d\theta\Biggr\rvert\leq\int_0^{2\pi/n}\biggl\lvert\frac{iRe^{i\theta}}{1+R^ne^{in\theta}}\biggr\rvert d\theta\leq\int_0^{2\pi/n}\biggl\lvert\frac{Re^{i\theta}}{R^ne^{in\theta}}\biggr\rvert d\theta = \frac{2\pi}{nR^{n-1}}
$$
Since our contour is a wedge, $n>1$. If $n=1$, we would be integrating over a circle instead.
$$
\lim_{R\to\infty}\int_0^{2\pi/n}\frac{iRe^{i\theta}}{1+R^ne^{in\theta}}d\theta\leq\lim_{R\to\infty}\frac{2\pi}{nR^{n-1}} = 0
$$
We are now only left with 
\begin{align}
(1-e^{2i\pi/n})\int_0^{\infty}\frac{dz}{1+z^n}&= \frac{-2i\pi e^{i\pi/n}}{n}\\
\int_0^{\infty}\frac{dx}{1+x^n} &= \frac{-2i\pi e^{i\pi/n}}{n(1-e^{2i\pi/n})}\\
&=\frac{\pi}{n}\csc(\pi/n)
\end{align}
A: On  $\gamma_3$ we have $z = e^{2 \pi i/n} x$ where $x$ goes from $R$ to $0$, so
$dz = e^{2\pi i/n} dx$ and $\int_{\gamma_3} \dfrac{1}{1+z^n} dz = e^{2\pi i/n} \int_R^0 \ldots$
