Evaluate $$ \int_{0}^{\infty}\frac{sin(ax)}{x}dx $$ using the cauchy-Goursat theorem.
I'm trying to understand this example given in my complex analysis lecture notes.
First, the initial problem is converted to an integral on the complex plane: $$ I = \int_{0}^{\infty}\frac{sin(ax)}{x}dx = \lim_{\delta\to 0} \int_{\delta}^{\infty}\frac{e^{iax}-e^{-iax}}{2ix} = \lim_{\delta\to 0}\int_{\mathbb{R}\setminus[-\delta,\delta]}\frac{e^{iaz}}{2iz}dz $$
Then, using this closed curve $\Gamma$ (Image link) and applying the Cauchy-Goursat theorem, we have:
$$ \oint_{\Gamma}\frac{e^{iaz}}{2iz}dz = \int_{\left(-R,R\right)\setminus\left[-\delta,\delta\right]}\frac{e^{iaz}}{2iz}dz + \int_{\pi}^{0}\frac{e^{ia\delta e^{i\theta}}}{2} d\theta + \int_{0}^{\pi}\frac{e^{iaRe^{i\theta}}}{2}d\theta = 0 $$
And the desired integral is just the first integral on the right side of this equation when $R\rightarrow \infty$ and $\delta \rightarrow 0$
The text just assumes that the limit can be taken inside the integralso we have
$$ \lim_{\delta\to 0}\int_{0}^{\pi}\frac{e^{ia\delta e^{i\theta}}}{2} d\theta = \frac{1}{2}\int_{0}^{\pi}d\theta = \frac{\pi}{2} $$
The other integral should vanish after the limit is taken
$$ \lim_{R\to\infty} \left[ \int_{0}^{\pi} e^{iaR e^{i\theta}} d\theta \right] = 0 $$
but I can't really see why this happens. Can anyone help me? Thanks in advance.