Complex analysis: Calculating $\int_{-\infty}^{\infty} \frac{\sin x}{x} dx$ by using $f(z) = \frac{e^{iz} - 1}{iz}$ [duplicate]

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.

Solution 1:

How do I get rid of the integral on the right?

Your last identity rewrites
$$\pi = \int_{-\infty}^{+\infty}\frac{\sin t}{t}dt -i \int_{-\infty}^{+\infty}\frac{\cos t -1}{t}dt,$$

then you may just take the real part of both sides.

Solution 2:

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