I want to calculate $\int _{x=0}^{\infty} \frac{\sin(x)}{x}$ with the function $f(z) = \frac{e^{iz}}{z}$.
I thought about using the closed path $\Gamma = \gamma _1 + \gamma _R + \gamma _2 + \gamma _{\epsilon}$, when:
$\gamma_1 (t) = t, t \in [i\epsilon, iR]$
$\gamma_R (t) = Re^{it}, t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
$\gamma_2 (t) = t, t \in [-iR, -i\epsilon]$
$\gamma_{\epsilon} (t) = \epsilon e^{it}, t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
I use the fact that $\frac{\sin(x)}{x}$ is an even function and has an anti derivative, so the integral on a closed path is zero.
I managed to show that $\int_{\gamma _{\epsilon}} f = -i\pi$ when $\epsilon \to 0$.
However I am struggling to show that $\int_{\gamma _R} f = 0$ when $R \to \infty$
Help would be appreciated
