Prove the integral $\int_{C_a} f(z) ~e^{iz}~\mathrm{d}z$ converges to $0$ when $a\to\infty$ Let $f(z)$ be a rational function without poles on the real axis. Assume that the degree of the numerator is strictly smaller than the degree of the denominator. Define $C_a$ be the arc $$\{ae^{it};t\in[0,\pi]\}.$$
where $a$ is a postive real number.
I want to know the integral
$$\int_{C_a} f(z)~e^{iz}~\mathrm{d}z$$
converges to $0$ when $a\to \infty$.

update: my friend told me this is Jordan's lemma.

[Has been proved to be wrong.] Alternatively, we need to prove
$$\int_{C_a} f(z)~\mathrm{d}z=\int_{C_a} \frac{p_n(z)}{q_m(z)}~\mathrm{d}z$$
converges to $0$ when $a\to \infty$, where $p_n$ and $q_m$ are the polynomials of degree $n$ and $m$ respectively with $n<m$.
It is derived from an exercise. Following the standard procedure here Example I, I stuck with the proof of the above statement.

Let $f$ be a rational function without poles on the real axis. Assume that the degree of the numerator is strictly smaller than the degree of the denominator. Show that 
  $$\int_{-\infty}^{\infty}f(x)e^{ix}~\mathrm{d} x=2\pi i\sum_{\mathrm{Im} z_k>0}\mathrm{res}_{z_k}(f(z)e^{iz})$$
  where $z_k$ denote the poles of $f$. The sum is over all poles in the upper half plane.

 A: 
Alternatively, we need to prove
  $$ \int_{C_a} f(z)~\mathrm{d}z=\int_{C_a} \frac{p_n(z)}{q_m(z)}~\mathrm{d}z $$
  converges to $0$ when $a\to\infty$.

No, this is not true if $n = m-1$. As a simple example, notice that
$$ \lim_{a\to\infty} \int_{C_a} \frac{dz}{z+i} = i\pi. $$
So you cannot drop out the factor $e^{iz}$. To show the convergence, let $a_0 > 0$ be large enough that every pole of $f(z)$ on $\mathbb{C}$ lies inside the circle $|z| = a_0$. Also pick $M > 0$ such that $|f(z)| \leq \frac{M}{|z|^{m-n}}$ for all $|z| > a_0$, Then for all $a > a_0$, we have
$$ \left| \int_{C} f(z)e^{iz} \, dz \right|
= \left| \int_{0}^{\pi} f(a e^{i\theta}) e^{ia e^{i\theta}} \, ia e^{i\theta} d\theta \right| 
\leq \frac{M}{a^{m-n-1}} \int_{0}^{\pi} e^{-a\sin\theta} \, d\theta. $$
Since $e^{-a\sin\theta} $ is uniformly bounded by $1$ and converges pointwise to $0$ for $\theta \in (0, \pi)$ as $a\to\infty$, dominated convergence theorem tells that the integral converges to $0$ as well. A more elementary solution is to utilize the bound $\sin\theta \geq \frac{2}{\pi}\theta$ for $0 \leq \theta\leq\frac{\pi}{2}$:
$$ \int_{0}^{\pi} e^{-a\sin\theta} \, d\theta
= 2\int_{0}^{\frac{\pi}{2}} e^{-a\sin\theta} \, d\theta
\leq 2\int_{0}^{\frac{\pi}{2}} e^{-\frac{2a}{\pi}\theta} \, d\theta
\leq \frac{\pi}{a}. $$
Therefore
$$ \left| \int_{C} f(z)e^{iz} \, dz \right| \leq \frac{\pi M}{a^{m-n}} \xrightarrow[a\to\infty]{} 0. $$
