Detailed proof of why integral over the upper semi-circle in $C$ of $\frac{e^{ix}}{x^2 + a^2}$ goes to $0$ as the radius goes to $\infty$? This is a follow up question to this question: Computing $\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}dx$ using residue calculus. For clarity, I'll reproduce the question here: calculate $\int_{-\infty}^\infty \frac{\cos x}{x^2 + a} \ dx$ using the residue theorem. The answer given lets $\gamma$ be the semi-circle centered at $0$ with radius $R$ in the upper half of the complex plane, and we can further let $\gamma_1$ be the arc of $\gamma$, and $\gamma_2$ be the part of $\gamma$ along the real line.
I'm working through the proof, which involves using the fact that $\cos x = \Re(e^{ix})$ to solve the problem of $\cos x$ not being bounded if $x$ is allowed to be imaginary. However, it is stated simply that the integral
$$\int_{\gamma_1}\frac{e^{ix}}{x^2 + a^2}\ dx \to 0\ \ \ \text{ as } \ \ \ R \to 0,$$
which I don't fully understand. I have the following bound on the integral:
$$\left|\int_{\gamma_1}\frac{e^{ix}}{x^2 + a^2}\ dx \right| \le \sup_{s \in \gamma_1}\left|\frac{e^{ix}}{x^2 + a^2}\right| \cdot \text{length}(\gamma_1). $$
But I am having a difficult time getting the RHS to go to $0$ as $R$ goes to $0$. Clearly, 
$$|e^{ix}| \le 1,$$
but then we are left with 
$$\left|\int_{\gamma_1}\frac{e^{ix}}{x^2 + a^2}\ dx \right| \le \sup_{s \in \gamma_1}\left|\frac{1}{x^2 + a^2}\right| \cdot \pi R. $$
Now, we have $x = z + iy$, where $-R \le z \le R$, and $0 \le y \le R$. But how can I find a lower bound on $|x^2 + a^2|$? I'm having a hard time visualizing what it should be. I've tried expanding out $x^2$, and I have that
$$|x^2+a^2| \ge |z^2 + 2izy + a^2| - |y^2|,$$
but can't get anywhere from there. It feels as if I should eventually get $|x^2 + a^2| \ge R^2+a^2,$ but I can't really prove this. 
I know this is probably a very silly question, but I seem to always be misunderstanding the details for proofs in complex analysis, and I want to make sure I understand every piece of this one.
 A: First a notational issue.  The variable in the integral over $\gamma_1$ should be $z$, not $x$, because $x$ in this context (to prevent ambiguity) needs to represent a real variable.  But on the arc $\gamma_1$, you'll necessarily encounter non-real values.
On $\gamma_1$ you can parameterize using $z = Re^{i\theta}$, where $0 \le \theta \le \pi$.  So actually the integral is:
$$ \int_{\gamma_1} \frac{e^{iz}}{z^2 + a^2} \, dz = \int_0^\pi \frac{e^{iRe^{i\theta}} Rie^{i\theta}}{R^2e^{2i\theta} + a^2} \, d\theta $$
Then with absolute values we get:
\begin{align}
  \left|\int_{\gamma_1} \frac{e^{iz}}{z^2 + a^2} \, dz\right| &= \left|\int_0^\pi \frac{e^{iRe^{i\theta}} Rie^{i\theta}}{R^2e^{2i\theta} + a^2} \, d\theta \right|\\[0.3cm]
    &\le \int_0^\pi \left|\frac{e^{iRe^{i\theta}} Rie^{i\theta}}{R^2e^{2i\theta} + a^2}\right| \, d\theta\\[0.3cm]
    &\le \frac{R\pi}{|R^2 - a^2|}\\[0.3cm]
    &\to 0 \text{ as } R \to +\infty
\end{align}
Where did the $|R^2 -a^2|$ in the denominator come from?  Just focusing on the denominator and using the reverse triangle inequality, we have
\begin{align}
  |R^2e^{2i\theta} + a^2| &= |R^2e^{2i\theta} - (-a^2)|\\
    &\ge \big||R^2e^{2i\theta}| - |-a^2|\big|\\
    &= |R^2 - a^2|
\end{align}
