How to get the inverse Laplace transform of an expression with irrational function term by residue theorem?

Question

$$F(s)=\frac{ e^{-A_2\sqrt{s}}}{s(\sqrt{s}+A_3)}$$

The difficulty is to deal with the term $(\sqrt{s}+A_3)$.

Answer 1

$$Res(F(s),0) = \lim_{s \rightarrow 0}s\frac{e^{-A_2 \sqrt{s}} e^{st}}{s(\sqrt{s}+A_3)} = \frac{1}{A_3}.$$ $$\gamma_3: -R \rightarrow 0, s=r e^{i\pi}; \gamma_5 =: 0 \rightarrow -R, s=r e^{-i\pi}.$$ $$\frac{1}{2\pi i}\int_{\gamma_3} \frac{e^{-A_2 \sqrt{s}} e^{st}ds}{s(\sqrt{s}+A_3)}=\frac{1}{2\pi i}\int_R^0 \frac{e^{-A_2 \sqrt{r}e^{i\pi/2}}\ e^{r e^{i\pi }\ t} e^{i\pi} dr}{re^{i\pi}(\sqrt{r}e^{i\pi/2}+A_3)} = \frac{1}{2\pi i}\int_0^R \frac{e^{-iA_2\sqrt{r}}e^{-rt}dr}{-r(i\sqrt{r}+A_3)},$$

$$\frac{1}{2\pi i}\int_{\gamma_5} \frac{e^{-A_2 \sqrt{s}} e^{st}ds}{s(\sqrt{s}+A_3)}=\frac{1}{2\pi i}\int_0^R \frac{e^{-A_2 \sqrt{r}e^{-i\pi/2}}\ e^{r e^{-i\pi}\ t} e^{-i\pi} dr}{re^{-i\pi}(\sqrt{r}e^{-i\pi/2}+A_3)} = \frac{1}{2\pi i}\int_0^R \frac{e^{iA_2\sqrt{r}}e^{-rt}dr}{-r(-i\sqrt{r}+A_3)}.$$

$$ILT(F(s)) = Res(F(s),0) - \frac{1}{2\pi i}\int_{\gamma_3} \frac{e^{-A_2 \sqrt{s}} e^{st}ds}{s(\sqrt{s}+A_3)} - \frac{1}{2\pi i}\int_{\gamma_5} \frac{e^{-A_2 \sqrt{s}} e^{st}ds}{s(\sqrt{s}+A_3)} = \frac{1}{A_3} - \frac{1}{\pi}\int_0^R \frac{A_3 \sin{A_2\sqrt{r}}+\sqrt{r}\cos{A_2\sqrt{r}}}{r(r+A_3^2)}e^{-rt}dr.$$