How to argue this integral over this semicircle vanishes as $R\to \infty$? I want to evaluate the inverse Laplace transform of the function $$F(s)=\frac{A(s)}{B(s)},\quad B(s)=(s-s_1)\cdots (s-s_n),\quad s_i\neq s_j,\forall i\ne j.$$
I want to use the inversion formula together with contour integration. My starting point is the inversion formula $$\mathcal{L}^{-1}[F(s)]=\dfrac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}F(s)ds,$$
where $\gamma \in \mathbb{R}$ is any number so that all singularities of $F(s)$ lie to its left in the complex plane.
In that case, assuming $A(s)$ has no singularities, the singularities are exactly $\{s_i\}$, so we take any $\gamma > \max \{\operatorname{Re}(s_i)\}$.
Next, I consider a contour $\Gamma$ made up of two pieces:


*

*$C_1(T)$: a vertical line $z(\lambda)=\gamma+i\lambda$ for $\lambda \in [-T,T]$. As $T\to \infty$ this one gives what we need.

*$C_2(T,R)$: a semicircle starting at $\gamma+iT$ and ending at $\gamma - iT$. 
We then have set $$g(z)=\dfrac{1}{2\pi i}e^{zt}F(z).$$ We then have
$$\int_{C_1(T)} g(z)dz = \oint_{\Gamma}g(z)dz-\int_{C_2(T,R)}g(z)dz.$$
The integral over the closed contour can be done with the residue theorem. My issue is the last integral. We must take $T,R\to \infty$ but for the Laplace transform I don't see how to proceed.
I mean, I have the intuition. Call $z = x+iy$, along $C_2(T,R)$ if $R\to \infty$, then $x\to -\infty$. Then $e^{zt} = e^{xt}e^{iyt}\to 0$ as $R\to \infty$, so it should make the integral decay as well. But that is not really rigorous.
So how can I proceed here and show the integral on the semicircle $C_2(T,R)$ vanishes as $T,R\to \infty$?
 A: You want to show that 
$$\left | \int_{C_2(T,R)} dz \, e^{z t} F(z) \right | \to 0$$
as $R \to \infty$.  On $C_2(T,R)$, $\left | e^{z t} \right | = e^{R t \cos{\theta}}$ where $\theta \in [\pi/2, 3 \pi/2]$.  Thus, the integral is bounded by
$$\begin{align} \max_{\theta \in [\pi/2, 3 \pi/2]} \left | F \left ( R e^{i \theta} \right ) \right | \int_{\pi/2}^{3 \pi/2} d\theta \,  e^{R t \cos{\theta}} & = \max_{\theta \in [\pi/2, 3 \pi/2]} \left | F \left ( R e^{i \theta} \right ) \right | \int_0^{\pi} d\theta \,  e^{-R t \sin{\theta}} \\ &= 2 \max_{\theta \in [\pi/2, 3 \pi/2]} \left | F \left ( R e^{i \theta} \right ) \right | \int_0^{\pi/2} d\theta \,  e^{-R t \sin{\theta}} \\ &\le 2 \max_{\theta \in [\pi/2, 3 \pi/2]} \left | F \left ( R e^{i \theta} \right ) \right | \int_0^{\pi/2} d\theta \,  e^{-2 R t \theta/ \pi} \\ &\le \frac{\pi}{R t} \max_{\theta \in [\pi/2, 3 \pi/2]} \left | F \left ( R e^{i \theta} \right ) \right | \end{align}$$
As long as $\max_{\theta \in [\pi/2, 3 \pi/2]} \left | F \left ( R e^{i \theta} \right ) \right | = o(R)$ as $R \to \infty$, then the integral over $C_2(T,R)$ vanishes in this limit.
