complex analysis - Calculate $int_{-infty}^{infty} frac{sin(z)}{z}dz$

$\Gamma_1$ is the line connecting $\epsilon$ to $R$.

$\Gamma_2$ is the counterclockwise upper semicircle with radius $R$.

$\Gamma_3$ is the line connecting $-R$ to $-\epsilon$.

$\Gamma_4$ is the clockwise upper semicircle with radius $\epsilon$.

Denote $\Gamma=\Gamma_1+\Gamma_2+\Gamma_3+\Gamma_4$.

I used Cauchy's integral theorem to show that

$\int_{\Gamma}^{} \frac{e^{iz}}{z}dz=0$.

Now I need to evaluate the different parts of the integral $\int_{\Gamma}^{} \frac{e^{iz}}{z}dz$ and show that $\int_{-\infty}^{\infty} \frac{\sin(z)}{z}dz=\pi$ using the fact that $\lim_{\epsilon\rightarrow 0} \int_{\Gamma_4}^{ } \frac{\sin(t)}{t}dt=-i\pi$.

So far I managed to show that $\int_{\Gamma_2}^{ } \frac{e^{iz}}{z}dz=0$.

So I am left with

$\int_{\Gamma_1}^{ } \frac{e^{iz}}{z}dz+\int_{\Gamma_3}^{ } \frac{e^{iz}}{z}dz=-\int_{\Gamma_2}^{ } \frac{e^{iz}}{z}dz \Rightarrow $

$\Rightarrow \int_{-R}^{-\epsilon} \frac{\cos(z)}{z}dz+i\int_{-R}^{-\epsilon} \frac{\sin(z)}{z}dz+\int_{\epsilon}^{R} \frac{\cos(z)}{z}dz+i\int_{\epsilon}^{R} \frac{\sin(z)}{z}dz=-\int_{\Gamma_2}^{ } \frac{\cos(z)}{z}dz-i\int_{\Gamma_2}^{ } \frac{\sin(z)}{z}dz$.

But I can't see how to evaluate these integrals. What am I missing here?

Answer

First note that by Jordan lemma:
$$
\lim_{R\to\infty}\int_{\Gamma_2}\frac{e^{iz}}{z}dz=0.
$$

Next
$$
\lim_{R\to\infty}\left[\int_{\Gamma_1}\frac{e^{iz}}{z}dz+\int_{\Gamma_3}\frac{e^{iz}}{z}dz\right]=

\int_{\epsilon}^\infty\frac{e^{ix}}{x}dx+\int_{-\infty}^{-\epsilon}\frac{e^{ix}}{x}dx=\int_{\epsilon}^\infty\frac{e^{ix}}{x}dx+\int_{\epsilon}^\infty\frac{e^{-ix}}{-x}dx\\
=2i\int_{\epsilon}^\infty\frac{\sin x}{x}dx.
$$
In the limit $\epsilon\rightarrow 0$ we have
$$
0=\oint\frac{e^{iz}}{z}dz=
\underbrace{-\pi}_{\Gamma_4} i+\underbrace{2i\int_{0}^\infty\frac{\sin x}{x}dx}_{\Gamma_1+\Gamma_3}\Rightarrow \int_0^\infty\frac{\sin x}{x}=\frac{\pi}{2}.
$$
Finally observing that $\frac{\sin x}{x}$ is an even function:
$$

\int_{-\infty}^\infty\frac{\sin x}{x}=2\int_0^\infty\frac{\sin x}{x}=\pi.
$$