1. ## Complex Analysis

I want to show that the integral $\int_{\gamma_R}\frac{dz}{z^{2n}+1} \to 0
$
as $R\to \infty$

,where $\gamma_R$ is the upper semi-circle from R to -R

For $n\geq 2$ this is easy to show. But how to show this for n=1?

2. Hmm.

$\displaystyle\int_{\gamma_{R}}\frac{1}{z^{2}+1}\,d z+\int_{-R}^{R}\frac{1}{x^{2}+1}\,dx=2\pi i\text{Res}[\frac{1}{z^{2}+1},i],$ right?

Thus,

$\displaystyle\int_{\gamma_{R}}\frac{1}{z^{2}+1}\,d z=2\pi i\text{Res}[\frac{1}{z^{2}+1},i]-\int_{-R}^{R}\frac{1}{x^{2}+1}\,dx.$

The RHS goes to zero as $R\to\infty.$

3. I don't see how this explains $\int_{\gamma_R}\frac{dz}{z^2+1}\to 0$

We can only state this with the knowledge that:

$\int_{-\infty}^{\infty}\frac{1}{x^2+1}dx = 2\pi i\mathrm{Res}[\frac{1}{z^2+1},i]$, wich frankly we don't.

The way it can be shown for $n\geq 2$ is by using $\gamma_R(t)=Re^{2\pi i t}, t\in [0,1]$

Then $\int_{\gamma_R}\frac{dz}{1+z^{2n}} \leq \sup_{t\in [0,1]}|\frac{2\pi iRe^{2\pi i t}}{(Re^{2\pi i t})^{2n}+1}|\cdot \pi R\to 0$ as $R\to\infty$

But for n=1, we can't tell this easy....at least i don't see how. Maybe with a different $\gamma_R$

4. Well, I think the approach works fine. But if you don't like it, you could also use an ML estimate. You have $L=\pi R,$ and

$\left|\dfrac{1}{R^{2}+1}\right|\le\left|\dfrac{1}{ R^{2}}\right|=:M.$ Therefore,

$\displaystyle\left|\int_{\gamma_{R}}\frac{1}{z^{2} +1}\,dz\right|\le ML=\frac{\pi R}{R^{2}}=\frac{\pi}{R}\to 0$ as $R\to\infty.$