I am solving the following inverse Laplace transform $$ \mathcal{L}^{-1}\left\{\sqrt{s}\right\}=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} \sqrt{s} e^{st}ds $$ I have chosen the following Bromwich contour
My problem lies in the ability to show that the the integral along the arc AB (also the integral along the arc EF), from $\frac{\pi}{2}$ to $\pi$ goes to zero as $R\rightarrow\infty$. So far what I have done is the following: Let $s=Re^{i\theta}$, then $$ \int_{ AB} \sqrt{s} e^{st}ds=\int_{\frac{\pi}{2}}^{\pi}R^{\frac{1}{2}}e^{\frac{i\theta}{2}} e^{xR({\cos\theta+i\sin\theta})}e^{i\theta}iRd\theta$$ Taking modulus in the integral we arrive at $$R^{\frac{3}{2}}\int_{\frac{\pi}{2}}^{\pi} e^{xR({\cos\theta})}d\theta $$ Making the substitution $\theta=\frac{1}{2}+\phi $ $$R^{\frac{3}{2}}\int_{0}^{\frac{\pi}{2}} e^{-xR{\sin\phi}}d\phi\leq R^{\frac{3}{2}}\int_{0}^{\frac{\pi}{2}} e^{-xR{\frac{2}{\pi}\phi}}d\phi $$ integrating the last integral I got $$ \frac{R^{\frac{1}{2}}\pi}{2x}\left[1-e^{-xR}\right]$$ which diverges as $R\rightarrow\infty$. Can someone please help me showing where I went wrong in this process. Thank you in advance.