First of all let's see the bromwich contour here. Let's denote $\Gamma$ as the curved portion $BJPKQLA$ of the Bromwich Contour. Then there's sufficient condition to make $$F(t)=\sum \text{ Residues of $e^{st}f(s)$ at poles of $f(s)$}\tag{1}$$ holds. Where $F(t)$ is the function of variable $t$ and $f(s)$ is the laplace transform of $F(t)$. My goal is to prove the following theorem.
Given theorem:
Theorem. If we can find constants $M>0$, $k>0$ such that on $\Gamma$ (where $s=Re^{i\theta}$), $$\left|f(s)\right| < \frac{M}{R^k}\tag{2}$$ then the integral around $\Gamma$ of $e^{st} f(s)$ approaches to zero as $R\to\infty$, i.e., $$\lim_{R\to \infty} \int_{\Gamma} e^{st}f(s)\, \Bbb ds =0\tag{3}$$
Questions:
- How to prove the Theorem?
- How to find constants $M$ and $k$, so that $(1)$ holds? Do we really need to find them since the $\color{red}{\text{red colored quantity}}$ on my attempt will vanish when $R \to \infty$?
Attempt:
IMO, i just need to show $(3)$ is true. I will suppose it first that on $\Gamma$ i have $$\left|f(s)\right| < \frac{M}{R^k}$$ And splitting the integral $(3)$ into 4 pieces ($BJ$, $JPK$, $KQL$, and $LA$) we get $$\int_{\Gamma} e^{st}f(s)\, \Bbb ds = \int_{\Gamma_1} e^{st}f(s)\, \Bbb ds + \int_{\Gamma_2} e^{st}f(s)\, \Bbb ds + \int_{\Gamma_3} e^{st}f(s)\, \Bbb ds + \int_{\Gamma_4} e^{st}f(s)\, \Bbb ds $$ Where $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, and $\Gamma_4$ represent $BJ$, $JPK$, $KQL$, and $LA$ respectively.
(To make it short, i'll just write for integral over $\Gamma_1$ or $BJ$.)
Along $\Gamma_1$ we have the following, since $s=Re^{i\theta}$, $\theta_0 \leq \theta \leq \pi/2$ $$I_1 = \int_{\Gamma_1} e^{st} f(s)\, \Bbb ds = \int_{\theta_0}^{\pi/2}e^{Re^{i\theta}t}f\left(Re^{i\theta}\right)iRe^{i\theta}\, \Bbb d\theta$$ Using upperbound estimation lemma we get $$\begin{align} \left|I_1\right| &= \int_{\theta_0}^{\pi/2} \left|e^{(R \cos{\theta})t}\right| \left|e^{i(R \sin{\theta})t}\right| \left|f\left(Re^{i\theta}\right)\right| \left|iRe^{i\theta}\right|\, \Bbb d\theta\\ &\leq \int_{\theta_0}^{\pi/2} e^{(R \cos{\theta})t} \left|f\left(Re^{i\theta}\right)\right| R\, \Bbb d\theta\\ &\leq \frac{M}{R^{k-1}} \int_{\theta_0}^{\pi/2} e^{(R \cos{\theta})t} \, \Bbb d\theta\\ &= \frac{M}{R^{k-1}} \int_0^{\phi_0} e^{(R \sin{\phi})t} \, \Bbb d\phi \end{align}$$
The last expression is obtained by setting $\theta= \pi/2 - \phi$ where $\phi_0 = \pi/2 -\theta_0 = \sin^{-1}{(\gamma/R)}$ (See the image link, $\gamma$ is a real number).
Since $\sin{\phi}<\sin{\phi_0}\leq \cos{\theta_0} = \gamma/R$, the last integral is less than or equal to
$$\begin{align} \frac{M}{R^{k-1}} \int_0^{\phi_0} e^{\gamma t} \, \Bbb d\phi &= \frac{Me^{\gamma t} \phi_0}{R^{k-1}} \\ &= \color{red}{\frac{Me^{\gamma t} \phi_0}{R^{k-1}} \sin^{-1}{\left(\frac{\gamma}{R}\right)}} \end{align}$$
As $R \to \infty$, the red colored quantity approaches to zero and the integral over $BJ$ is zero. Seems we ignore the constants $M$ and $k$? Somehow, i'm not sure this proves the theorem.
This attempt is adapted from "Schaum's outlines Laplace Transforms pg. 203"
Hope you can help me. As always, thanks in advance!