Complex analysis: Calculating $\int_{-\infty}^{\infty} \frac{\sin x}{x} dx$ by using $f(z) = \frac{e^{iz} - 1}{iz}$ So I integrate the holomorphic function $\frac{e^{iz} -1}{iz}$ over the half disk in the upper half plane, let me name it $\Gamma$. By using Cauchy's theorem, it equals 0.
$$0 = \int_{\Gamma} \frac{e^{iz} -1}{iz} = \int_{0}^{\pi}(e^{iRe^{i\theta}}-1)d\theta + \int_{-R}^{R} \frac{e^{it}-1}{it}dt$$
($Re^{[0, \pi]}$ and $t: [-R, R]$ being the respective parametrizations.) 
By splitting the left integral in two, I get $\int_{0}^{\pi} e^{iRe^{i\theta}}d\theta \rightarrow 0$ by dominated convergence, and $\int_{0}^{\pi} -1d\theta = -\pi$. By making $R \rightarrow \infty$, I'm left with
$$\pi = \int_{-\infty}^{\infty} \frac{e^{it} - 1}{it}dt = \int_{-\infty}^{\infty}\frac{\sin t}{t}dt + \int_{-\infty}^{\infty}\frac{\cos t -1}{it}dt.$$
How do I get rid of the integral on the right? I welcome all feedback. 

$\frac{\cos(t)-1}{it}$ is an odd function so its integral over any $[-R,R]$ is zero.
