Theorem: Jordan's Lemma

Let $r > 0$ be a real number.
Let:
:$C_r = \set {r e^{i \theta}: 0 \le \theta \le \pi}$
Let $g : C_r \to \C$ be a continuous function.
Define $f : C_r \to \C$ by: 
:$\map f z = e^{i a z} \map g z$
for each $z \in C_r$, for some real number $a > 0$. 
Then:
:$\ds \size {\int_{C_r} \map f z \rd z} \le \frac \pi a \paren {\max_{0 \mathop \le \theta \mathop \le \pi} \size {\map g {r e^{i \theta} } } }$

Proof:
We have: 
{{begin-eqn}}
{{eqn | l = \size {\int_{C_r} \map f z \rd z}
      | r = \size {\int_{C_r} e^{i a z} \map g z \rd z} 
}}
{{eqn | r = \size {i r} \size {\int_0^\pi e^{i a r e^{i \theta} } \map g {r e^{i \theta} } e^{i \theta} \rd \theta}
      | c = {{Defof|Complex Contour Integral}}
}}
{{eqn | r = \size {i r} \size {\int_0^\pi e^{i a r \paren {i \sin \theta + \cos \theta} } \map g {r e^{i \theta} } e^{i \theta} \rd \theta}
      | c = Euler's Formula
}}
{{eqn| r = r \size {\int_0^\pi e^{a r \paren {i \cos \theta - \sin \theta} } \map g {r e^{i \theta} } e^{i \theta} \rd \theta}
      | c = $i^2 = -1$, $\size i = 1$
}}
{{eqn | o = \le
      | r = r \int_0^\pi \size {e^{i a r \cos \theta} } \size {e^{-a r \sin \theta} } \size {\map g {r e^{i \theta} } } \size {e^{i \theta} } \rd \theta
      | c = Modulus of Complex Integral
}}
{{eqn | r = r \int_0^\pi e^{-a r \sin \theta} \size {\map g {r e^{i \theta} } } \rd \theta
      | c = $\size {e^{i \theta} } = 1$ for real $\theta$
}}
{{eqn | o = \le 
      | r = r \int_0^\pi e^{-a r \sin \theta} \max_{0 \le \theta \le \pi} \size {\map g {r e^{i \theta} } } \rd \theta
      | c = {{Defof|Supremum of Real-Valued Function}}
}}
{{end-eqn}}
From Definite Integral of Constant Multiple of Real Function, we can write:
:$\ds r \int_0^\pi e^{-a r \sin \theta} \max_{0 \le \theta \le \pi} \size {\map g {r e^{i \theta} } } \rd \theta = r \paren {\max_{0 \le \theta \le \pi} \size {\map g {r e^{i \theta} } } } \int_0^\pi e^{-a r \sin \theta} \rd \theta$
We now focus on: 
:$\ds \int_0^\pi e^{-a r \sin \theta} \rd \theta$
We have: 
{{begin-eqn}}
{{eqn | l = \int_0^\pi e^{-a r \sin \theta} \rd \theta
      | r = \int_0^{\pi/2} e^{-a r \sin \theta} \rd \theta + \int_{\pi/2}^\pi e^{-a r \sin \theta} \rd \theta
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \int_0^{\pi/2} e^{-a r \sin \theta} \rd \theta - \int_{\pi/2}^0 e^{-a r \map \sin {\pi - \theta} } \rd \theta
      | c = substituting $\theta \mapsto \pi - \theta$]]
}}
{{eqn | r = \int_0^{\pi/2} e^{-a r \sin \theta} \rd \theta + \int_0^{\pi/2} e^{-a r \sin \theta} \rd \theta
      | c = Sine of Angle plus Integer Multiple of Pi, Sine Function is Odd
}}
{{eqn | r = 2 \int_0^{\pi/2} e^{-a r \sin \theta} \rd \theta
}}
{{end-eqn}}
Note that we also have, from Jordan's Inequality: 
:$\sin \theta \ge \dfrac {2 \theta} \pi$
We therefore have: 
{{begin-eqn}}
{{eqn | l = 2 \int_0^{\pi/2} e^{-a r \sin \theta} \rd \theta
      | o = \le
      | r = 2 \int_0^\pi e^{-\paren {2 a r \theta} / \pi} \rd \theta
}}
{{eqn | r = 2 \intlimits {-\frac {\pi e^{\paren {2 a r \theta} / \pi} } {2 a r} } 0 \pi
      | c = Primitive of Exponential of a x, Fundamental Theorem of Calculus
}}
{{eqn | r = \frac {\pi \paren {1 - e^{-2 a r} } } {a r}
}}
{{eqn | o = \le
      | r = \frac \pi {a r}
      | c = $e^{-a r} < 1$ for positive $a, r$
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \size {\int_{C_r} \map f z \rd z}
      | o = \le
      | r = r \paren {\max_{0 \le \theta \le \pi} \size {\map g {r e^{i \theta} } } } \int_0^\pi e^{-a r \sin \theta} \rd \theta
}}
{{eqn | o = \le
      | r = \frac \pi {a r} \times r \times \paren {\max_{0 \le \theta \le \pi} \size {\map g {r e^{i \theta} } } }
}}
{{eqn | r = \frac \pi a \paren {\max_{0 \le \theta \le \pi} \size {\map g {r e^{i \theta} } } }
}}
{{end-eqn}}
{{qed}}