complex analysis - Compute the Integral via Residue Theorem

My goal is to compute $$I=\int_{0}^{+∞}\frac{\cos{ax}}{1+x^2}dx$$ where $a>0$.

$$I=\frac{1}{2}\int_{-∞}^{+∞}\frac{\cos{ax}}{1+x^2}dx=\frac{1}{2}Re\bigg(\int_{-∞}^{+∞}\frac{e^{iax}}{1+x^2}dx\bigg)$$ .

Let $f(z)=\frac{e^{iaz}}{1+z^2}$.

By Residue Theorem, $\int_{-R}^{R}\frac{e^{iax}}{1+x^2}dx+\int_{\gamma_R}\frac{e^{iaz}}{1+z^2}dz=2\pi Res(f,i)=\frac{e^{-a}}{2i}$, where $\gamma_R$ denotes the upper semi-circle centered at $O$ with radius $R$.

As $R$—>$+∞$,

$\int_{-R}^{R}\frac{e^{iax}}{1+x^2}dx$ —> $\int_{-∞}^{+∞}\frac{e^{iax}}{1+x^2}dx$

Now, I am stuck on how to prove $\int_{\gamma_R}\frac{e^{iaz}}{1+z^2}dz$ goes to $0$ as $R$ goes to infinity.

Anyone know how to do it? Many thanks.

Answer

Note that for $z=R(\cos(t)+i\sin(t))$ with $R>1$ and $t\in [0,\pi]$
$$\left|\frac{e^{iaz}}{1+z^2}\right|=\frac{e^{-aR\sin(t)}}{R^2-1} \leq \frac{1}{R^2-1}.$$
Hence, as $R\to +\infty$,
$$\left|\int_{\gamma_R}\frac{e^{iaz}}{1+z^2}dz\right|\leq \frac{|\gamma_R|}{R^2-1}=\frac{\pi R}{R^2-1}\to 0.$$
P.S. This is a particular case of the Jordan Lemma.