In this post i will prove:

Jordan’s Lemma: Let $\alpha \in (0, \infty)$, $R \in (0, \infty)$, $\gamma_R : [0, \pi] \to \mathbb{C}$, $\gamma_R(t):= R \exp (it)$ and $f: \gamma_R ([0, \pi]) \to \mathbb{C}$ continuous. Then \begin{equation} \bigg| \int_{\gamma_R} \exp ( \alpha i z) f (z) dz \bigg| \leq M_R \frac{\pi}{\alpha}, \end{equation} where $M_R:= \max_{ t \in [0, \pi]} |f(R e^{it})|$.

Proof: By definition of the curve integral: \begin{equation} \bigg| \int_{\gamma_R} \exp ( \alpha i z) f (z) dz \bigg| =\bigg| \int_0^{\pi} \exp ( \alpha i R e^{it}) f (R e^{it}) R i e^{it} dt \bigg|. \end{equation} Moving the absolute value into the integral: \begin{equation} \bigg| \int_0^{\pi} \exp ( \alpha i R e^{it}) f (R e^{it}) R i e^{it} dt \bigg| \leq \int_0^{\pi} |\exp ( \alpha i R e^{it})| |f (R e^{it})| R |ie^{it}| dt \end{equation} and therefore using the monotonicity of the integral: \begin{equation} \bigg| \int_{\gamma_R} \exp ( \alpha i z) f (z) dz \bigg| \leq M_R R \int_0^{\pi} |\exp ( \alpha i R e^{it})| dt. \end{equation} It is left to estimate the integral. In general: $\forall z := x+iy \in \mathbb{C}: |e^z | = e^{x}$ and therefore for all $t \in [0,\pi]$: \begin{equation} |\exp ( \alpha i R e^{it})| = \exp (- \alpha R \sin (t) ), \end{equation} where $e^{it} = \cos (t) + i \sin (t)$ was used. Therefore \begin{equation} \int_0^{\pi} |\exp ( \alpha i R e^{it})| dt = \int_0^\pi \exp (- \alpha R \sin (t) )dt \end{equation} The idea now is to show that \begin{equation} \int_0^\pi \exp (- \alpha R \sin (t) )dt = 2\int_0^\pi \exp (- \alpha R \sin (t) )dt \end{equation} and to then use that $\sin$ is concave on $[0,\pi/2]$ together with the monotonicity of the exponential function to estimate the integral from above. Define $\varphi : [0,\pi] \to [0,\pi]$, $\varphi (t):= (\pi -t)$. Then $\varphi^\prime =-1$ and $\forall t \in [0,\pi]$: $\sin (\varphi (t) ) = \sin(t)$ (follows from addition theorem). Using substitution: \begin{equation} \int_0^{\pi/2} \varphi^\prime (t) \exp (- \alpha R \sin (\varphi (t)) )dt = \int_{\varphi (0)}^{\varphi(\pi)}\exp (- \alpha R \sin (t) )dt \end{equation} and so \begin{equation} - \int_0^{\pi/2} \exp (- \alpha R \sin (t))dt = \int_\pi^{\pi/2} \exp (- \alpha R \sin (t) )dt \end{equation} From here it follows from the basic integration rules that \begin{equation} \int_0^{\pi} \exp (- \alpha R \sin (t))dt = 2 \int_0^{\pi/2} \exp (- \alpha R \sin (t))dt. \end{equation} Now $\sin$ is concave on $[0,\pi/2]$ (follows from the derivative characterisation). Therefore by definition of concavity $\forall x_1, x_2 \in \mathbb [0,\pi/2]$ and $\lambda \in [0,1]$: \begin{equation} \sin (x_1 \lambda + x_2 (1-\lambda)) \geq \lambda \sin (x_1) + (1-\lambda) \sin (x_2). \end{equation} Let $y \in [0, \pi/2]$. Choosing $x_2 =0$, $x_1 = \pi/2$ and $\lambda = y /(\pi /2)$ in the above: \begin{equation} \sin ( y) \geq \frac{y}{\pi /2} \Rightarrow - \sin ( y) \leq -\frac{y}{\pi /2}. \end{equation} Since $\exp$ (on $\mathbb{R}$) and the integral are monotone: \begin{equation} \int_0^{\pi/2} \exp (- \alpha R \sin (t))dt \leq \int_0^{\pi/2} \exp (- \alpha R \frac{t}{\pi /2} )dt \end{equation} and \begin{equation} \int_0^{\pi/2} \exp (- \alpha R \frac{t}{\pi /2} )dt = \frac{-\pi}{2\alpha R} ( \exp (- \alpha R) - 1) \leq \frac{\pi}{2\alpha R}, \end{equation} since $\exp \geq 0$ on $\mathbb{R}$. Putting everything together: \begin{equation} \bigg| \int_{\gamma_R} \exp ( \alpha i z) f (z) dz \bigg| \leq M_R R \int_0^{\pi} |\exp ( \alpha i R e^{it})| dt \leq M_R R 2 \frac{\pi}{2\alpha R} = M_R \frac{\pi}{\alpha}. \end{equation} Which ends the proof. $\square$