Complex Analysis: Show that $\int_{C_\epsilon} \frac{1}{(z-x_0)^{2n+1}}dz = 0$

Question

We have to show that:

$$\int_{C_\epsilon} \frac{1}{(z-x_0)^{2n+1}}dz = 0$$

Where $x_0 \in \mathbb{R}$, $n$ is a positive integer and $C_\epsilon$ is the path that follows counterclockwise, in the complex plane, a semi circumference centred at $x_0$, with $\epsilon$ radius and non-negative imaginary part.

Attempt: I don't know if this integral can be solved by using Cauchy's integral formula. I've tried it by a series expansion of the function inside the integral but I get a very complicated expression that I guess it's not the right way for solving this problem.

Could it be done, maybe, by using the Residue theorem? Could you give me any hint?

Thanks in advance!

Answer 1

Suppose $U$ is an open subset of the complex plane $\mathbb C, f\colon U → \mathbb C$ is a function, and $\mathcal{C}\subset U$ is a curve of finite length, parametrized by $\gamma : [a,b] → \mathcal C$. If the parametrization $\gamma$ is continuously differentiable, the line integral can be evaluated as an integral of a function of a real variable: $$\int_\mathcal{C} f(z)\,dz = \int_a^b f(\gamma(t)) \gamma'(t)\,\mathrm{d}t.$$

To evaluate your contour, we have to find it's parameterization first. Note that the function $e^{it}, t\in[-\pi,\pi]$ graphs the unit circle in the complex plane. To convert it into a semic circle, we can simply let $t\in[0,\pi]$. If we multiply this function by some real number, the semicircle get's bigger, so the factor measures it's radius. Now $\varepsilon e^{it}, t\in[0,\pi]$ is a semicircle of radius $\varepsilon$. If we want to shift the center of the semicircle, we simply add a complex number. Now we may write that $\gamma(t)=x_0+\varepsilon e^{it}, t\in[0,\pi]$, so

$$\int_{C_\varepsilon} \frac{1}{(z-x_0)^{2n+1}}\,\mathrm{d}z=\int_{0}^{\pi} \frac{1}{(x_0+\varepsilon e^{it}-x_0)^{2n+1}}\cdot\left[i\varepsilon e^{it}\right]\,\mathrm{d}t.$$

Note that the factor in the square brackets is $\gamma'(t)$. We may eliminate $x_0$ and take $\varepsilon$ out of the integral to get
$$\frac{i}{\varepsilon^{2n}}\int_{0}^{\pi} \frac{1}{(e^{it})^{2n+1}}e^{it}\,\mathrm{d}t=\frac{i}{\varepsilon^{2n}}\int_{0}^{\pi} e^{-2nit}\,\mathrm{d}t.$$ Since $e^{-2nit}$ has a holomorphic antiderivative, we can apply the FTC which yields $$\frac{i}{\varepsilon^{2n}}\left[e^{-2ni\cdot\pi}-e^{-2ni\cdot0}\right]=\frac{i}{\varepsilon^{2n}}\left[1-1\right]=0.$$

The last line, $e^{-2ni\cdot\pi}=1$, holds for $n\in\mathbb{Z}$.

Answer 2

Just calculate it explicitly: $\int_0^{\pi} \frac 1 {\epsilon e^{i(2n+1)\theta}} \epsilon i e^{i\theta} d\theta=0$ since $\int_0^{\pi} e^{-2ni\theta}d\theta=0$.

[The path is $x_0+\epsilon e^{i\theta}, 0<\theta <\pi$].

Answer 3

$F(z)=-\frac{1}{2n}\frac{1}{(z-z_0)^{2n}}$ is a primitive of the given function, and integrals over functions with a primitive are easy to calculate: $\int_C f(z)\mathrm dz=F(b)-F(a)$, where $a$ and $b$ are the start and end point, respectively. In this case that's

$$F(z_0+\varepsilon)-F(z_0-\varepsilon)=-\frac{1}{2n}\left(\frac{1}{\varepsilon^{2n}}-\frac{1}{(-\varepsilon)^{2n}}\right)=0.$$