I've been using this example $$\mathcal{L}^{-1} \{\frac{3}{(s^2+9)^2}\}(t)$$ Actually calculating the inverse is fine, but I got stuck trying to prove that the circular part of the Bromwich contour goes to $0$. If the angle to the top of the circular path from the x-axis is $\alpha$, then the whole path rotates $2(\pi-\alpha)$ :
$$\int_{0}^{2(\pi-\alpha)}{\frac{3e^{st}}{(s^2+9)^2}\,ds}$$
$s=Re^{i\theta},ds=iRe^{i\theta}d\theta$
$$3iR\int_{0}^{2(\pi-\alpha)}{\frac{e^{Rte^{i\theta}}}{(R^2e^{i2\theta}+9)^2}e^{i\theta}\, d\theta}$$
Using $|\int_{a}^{b}{f(x)\, dx}|\leq\int_{a}^{b}{|f(x)|\, dx}$ and the reverse triangle inequality on the denominator :
$$\begin{align}\left|3iR\int_{0}^{2(\pi-\alpha)}{\frac{e^{Rte^{i\theta}}}{(R^2e^{i2\theta}+9)^2}e^{i\theta}\,d\theta}\right|&\leq 3iR\int_{0}^{2(\pi-\alpha)}{\left|e^{Rte^{i\theta}}\right|\frac{1}{\left|R^2e^{i2\theta}+9\right|^2}\left|e^{i\theta}\right|\, d\theta}\\&\leq3iR\int_{0}^{2(\pi-\alpha)}{\left|e^{Rt\cos(\theta)}\right|\left|e^{iRt\sin(\theta)}\right|\frac{1}{|R^2-9|^2}\, d\theta}\\&\leq\frac{3iR}{|R^2-9|}\int_{0}^{2(\pi-\alpha)}{\left|e^{Rt\cos(\theta)}\right|\, d\theta} \end{align}$$
How can I continue this to show that the expression goes to $0$ as $R\to\infty$? I'm worried that the use of the reverse triangle inequality and the bounds of integration are incorrect.
Edit : The contour looks like the image on below, which is from here
The Bromwich contour lies on the straight-line path $s=c+iy$, $c>0$ and $y\in (-\infty,\infty)$. So, the semicircular contour $C_R$ is defined by $s=Re^{i\theta}$ where $\theta\in [\pi/2-\arcsin(c/R) ,3\pi/2+\arcsin(c/R)]$. Thus, we have for $R>3$
$$\left|\int_{\pi/2-\arcsin(c/R)}^{3\pi/2+\arcsin(c/R)}\frac{3e^{Rte^{i\theta}}}{((R^2e^{i2\theta})^2+9)^2}\,iRe^{i\theta}\,d\theta\right|\le \frac{3R}{(R^2-9)}\int_{\pi/2-\arcsin(c/R)}^{3\pi/2+\arcsin(c/R)}e^{Rt\cos(\theta)}\,d\theta\tag1$$
The integral on the right-hand side of $(1)$ can be estimated as follows.
$$\begin{align} \int_{\pi/2-\arcsin(c/R)}^{3\pi/2+\arcsin(c/R)}e^{Rt\cos(\theta)}\,d\theta&=2\int_{\pi/2-\arcsin(c/R)}^{\pi}e^{R\cos(\theta)}\,d\theta\\\\ &=2\int_{-\arcsin(c/R)}^{\pi/2}e^{-Rt\sin(\theta)}\,d\theta\\\\ &=2\int_0^{\arcsin(c/R)}e^{Rt\sin(\theta)}\,d\theta+2\int_0^{\pi/2}e^{-Rt\sin(\theta)}\,d\theta\\\\ &\le 2\arcsin(c/R)e^{ct}+2\int_0^{\pi/2}e^{-Rt(2\theta/\pi)}\,d\theta\\\\ &=2\arcsin(c/R)e^{ct}+\frac{\pi(1-e^{-Rt})}{Rt} \end{align}$$
which clearly goes to $0$ as $R$ tends to infinity.