Jordan Lemma - Complex Integral How does the all of the LHS equal zero?
$$\lim_{R\to\infty} \int_{H_R} \frac{e^{imz}\,dz}{a^2 + z^2}=0$$
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 A: Why not directly with the ML inequality, for $\;m>0\;$? :
$$\left|\int_{H_R}\frac{e^{imz}}{z^2+a^2}dz\right|\le\ell(H_R)\max_{z\in H_R}\frac{|e{^{imz}|}}{|z^2+a^2|}\le\pi R\frac{e^{-m\,\text{Im}\,z}}{R^2-a^2}\xrightarrow[R\to\infty]{}0$$
since Im$\,z>0\;$ for $\;z\in H_R\;$
A: Let $m>0$
Proposition : 
Assuming $H_R$ denotes the arc spanning the real axis of radius $R$, the limit of the integral is $0$ as $R$ tends to infinity because $|zf(z)|$ goes to $0$ as $|z|$ goes to infinity, where $f$ denotes the integrand.
Proof :
Intuitively speaking, $|z|$ is the distance of the point $(x,y)$ to the origin $(0,0)$ in the complex plane.
In a more formal way, if $z=x+iy$, then $|z|=\sqrt{x^2+y^2}$.
As the semi-circle radius grows to infinity, the distance of one of its points relative to the origin grows to infinity to. Indeed, the semi-circle of radius R in the upper plane is just :
{$z=Re^{i\theta},\theta \in ]0,\pi[$}
and $|Re^{i\theta}|=R$.
So as $R$ grows and tends to infinity, $|z|$ tends to infinity too.
We are interested in
$$\lim_{R\to\infty} \int_{H_R} \frac{e^{imz}\,dz}{a^2 + z^2}$$
First, notice that 
$$|\int_{H_R} \frac{e^{imz}\,dz}{a^2 + z^2}|=|\int_{H_R} f(z)dz|=|\int_{H_R} zf(z)\frac{dz}{z}|$$
Since $z=Re^{i\theta}$, with R constant because we are integrating $z$ over a semi-circle, $dz=Rd(e^{i\theta})=Rie^{i\theta}d\theta=izd\theta$
So
$$|\int_{H_R} zf(z)\frac{dz}{z}|=|\int_{0}^{\pi} zf(z)id\theta|=|\int_{0}^{\pi} zf(z)d\theta|\times|i|=|\int_{0}^{\pi} zf(z)d\theta|\leq \int_{0}^{\pi} |zf(z)|d\theta \leq \pi sup(|zf(z)|)$$
However, $|zf(z)|=\frac{|z|}{|z^2+a^2|} \times |e^{imz}|$
We have
$lim _{|z| \rightarrow \infty}\frac{|z|}{|z^2+a^2|}=0$ because the denominator clearly dominates the numerator
and if $z=x+iy$, we must have $y>0$ because the semi-circle is located in the upper plane, so :
$|e^{imz}|=|e^{imx-my}|=e^{-my} \times |e^{imx}|=e^{-my}$
so $lim_{|z| \rightarrow +\infty}|e^{imz}|=lim_{y \rightarrow +\infty}e^{-my}=0$
Note that this trick wouldn't have worked in the lower half plane, since $y$ would have been negative and the minus sign in the exponential would become a plus sign, making the limit explode to infinity.
As a result, 
$$\lim_{R\to\infty} |\int_{H_R} \frac{e^{imz}\,dz}{a^2 + z^2}| \leq \pi\times lim_{|z| \rightarrow +\infty} sup(|zf(z)|)=0$$
So $$\lim_{R\to\infty} \int_{H_R} \frac{e^{imz}\,dz}{a^2 + z^2}=0$$
