Complex Analysis-Path Integral

[notmuch]

Complex Analysis--Path Integral

Homework Statement

Let I(r) = Int(e^(iz)/z) over the "top half" of the circle of radius r, centered at the origin. Show that lim {r -> infty} I(r) = 0.

Homework Equations

All given.

The Attempt at a Solution

I was thinking of using the inequality |Int(e^(iz)/z)| <= Int(e^(-r*sin t)) from 0 to pi. I want to show that the right hand side of the inequality goes to zero as r -> infty. If so, then the problem should be solved. Thanks.

 

[notmuch]

A re-statement of the problem and my work (so it's easier to read):

Let $$\gamma(t) = re^{it}$$ for $$t \in [0, \pi]$$.
Evaluate:

$$\lim_{r \to \infty}\int_{\gamma}{\frac{e^{iz}}{z}}$$

So far I have:

$$\int_{\gamma}{\frac{e^{iz}}{z}} = \int_{0}^{\pi}{e^{-r \sin t}(\cos(r\cos t) + i\sin(r\cos t)}dt$$

So,
$$|\int_{\gamma}{\frac{e^{iz}}{z}}| \leq \int_{\gamma}{|\frac{e^{iz}}{z}|} = \int_{0}^{\pi}{e^{-r\sin t}$$

Am I on the right track? Anyone have a suggestion for showing that $$\lim_{r \to \infty} \int_{0}^{\pi}{e^{-r\sin t} = 0$$?

Thanks!