Showing contour integral $\int_{C_R}\frac{e^{st}}{\sqrt{s}}\,ds$ goes to $0$ I was working on a question similar to this: Special Laplace Inversion, and reading through the solution I couldn't figure out how to justify that the integral along the circular arcs goes to $0$ as $R\rightarrow\infty$ and $r\rightarrow0$. I tried to parameterise as $p=c+Re^{i\theta}$, but then in the integral $\int_{c-i\infty}^{c+i\infty}e^{pt}p^{-\frac{1}{2}} \text{d}p$, there ends up being a term $\log|c+Re^{i\theta}|$, which doesnt really simplify. Is there a better approach to this?
 A: We cut the plane from $0$ to $-\infty$ along the non-positive real axis.  The parameterization of the circular arc is $s=Re^{i\theta}$, where $-\pi \theta\le \pi$.   Then, we see that
Then, we have
$$\begin{align}
\lim_{R\to \infty}\int_{C_R}\frac{e^{st}}{\sqrt{s}}\,ds&=\lim_{R\to \infty}\int_{\pi/2-\arctan(a/R)}^{\pi^-}\frac{e^{Re^{i\theta}t}}{\sqrt{R}e^{i\theta/2}}\,iRe^{i\theta}\,d\theta\\\\
&+\lim_{R\to \infty}\int_{-\pi^+}^{-\pi/2+\arctan(a/R)}\frac{e^{Re^{i\theta}t}}{\sqrt{R}e^{i\theta/2}}\,iRe^{i\theta}\,d\theta\\\\
&=\lim_{R\to \infty}2i\sqrt{R} \int_{\pi/2-\arctan(a/R)}^{\pi^-} e^{R\cos(\theta t)}\cos\left(R\sin(\theta)t+\theta/2\right)\,d\theta\tag1
\end{align}$$

Next, we have the estimates for the integral on the right-hand side of $(1)$.
$$\begin{align}
\left|2i\sqrt{R}\int_{\pi/2-\arctan(a/R)}^{\pi^-}e^{R\cos(\theta)t}e^{i(R\sin(\theta)t+\theta/2)}\,d\theta\right|&\le 2\sqrt{R}\int_{\pi/2-\arctan(a/R)}^{\pi^-}e^{R\cos(\theta)t}\,d\theta\\\\
&=2\sqrt{R}\int_{-\arctan(a/R)}^{\pi/2} e^{-R\sin(\theta)t}\,d\theta\\\\
&\le 2\sqrt{R}\int_{-\arctan(a/R)}^0e^{-R\sin(\theta)t}\,d\theta\\\\
&+2\sqrt{R}\int_{0}^{\pi/2}e^{-R\sin(\theta)t}\,d\theta\\\\
&\le2\sqrt{R}\int_{-\arctan(a/R)}^0e^{-R\theta t}\,d\theta\\\\
&+2\sqrt{R}\int_{0}^{\pi/2}e^{-R2\theta t/\pi}\,d\theta\\\\
&=\frac{2\left(e^{R\arctan(a/R)t}-1\right)}{\sqrt{R}t}\\\\
&+\frac{\pi\left(1-e^{-Rt}\right)}{\sqrt{R}t}\tag 2
\end{align}$$

Note that $\lim_{R\to \infty}R\arctan(a/R)=a$.  
Therefore, as $R\to \infty$, the terms on the right-hand side of $(2)$ approach $0$ and we assert that 

$$\lim_{R\to \infty}\int_{C_R}\frac{e^{st}}{\sqrt{s}}\,ds=0$$

as was to be shown.
