$\lim_{r\to +\infty}\int_{\partial D_r\left (0\right )}\frac{e^{iz}}{z}dz=0$

Question

Let $D_r\left (0\right )$ the disc of center $r$ with center $0$ in $\mathbb{C}$ and let $\partial D_r\left (0\right )$ be its boundary.

Prove that $$\lim_{r\to +\infty}\int_{\partial D_r\left (0\right )}\frac{e^{iz}}{z}dz=0$$

I tried to solve this problem using estimation lemma, but the upper bound I found is $2\pi e^r$ which does not tend to zero.

It is my first approach to integrals over curves defined on the complex plane, therefore I do not have much theory, but I am supposed to be able to solve this problem. Any help?

Answer 1

There is probably an error in the definition of the curve of integration, and it is assumed you only transverse the circle when $0\leqslant t\leqslant \pi$. First compute the integral explicitly.
$$ \int_0^{\pi }\frac{ \exp( ir e^{it}) ire^{it}}{re^{it}}dt = i\int_0^{\pi } \exp(ir e^{it})dt = i\int_0^{\pi} \exp(ir\cos t)\exp(-r\sin t)dt $$

It follows that the integral, in absolute value, is at most

$$I(r) = \int_0^{\pi}\exp(-r\sin t)dt $$

Now $\sin t \geqslant \dfrac{2t}\pi$ in $[0,\pi/2]$, so the claim follows for the integral in such subinterval because $$\int_0^{\pi/2} \exp({-2tr/\pi})dt \to 0,$$ and the integral over the other subinterval is symmetric to this.

Answer 2

Using cauchys integral formula: \begin{align*} f(z) = \frac{1}{2\pi i} \oint_\gamma \frac{f(w)}{w-z}dw, \end{align*} setting $f(z)=\exp(iz)$ we have that:

\begin{align*} \frac{1}{2\pi i} \oint_\gamma \frac{\exp(iw)}{w-0}dw = \exp(i0)=1. \end{align*}

Therefore we have that: \begin{align*} \oint_{\partial D_r(0)} \frac{\exp(iw)}{w}dw = 2\pi i. \end{align*}

So you limit is $2\pi i$ and not 0.