Calculating the inverse Laplace transform of $F_{5}(s)=\frac{-A_{10} e^{-A_{7} \sqrt s }}{(\sqrt s+A_8) \sqrt{S+\theta_c}(S-A_4)}$ by residue theorem

Question

I have calculated the ILT by residue theorem, but the result is wrong. Would you like to check the calculation. Thanks.

Answer 1

$$Res(F_{5}(s),A_{4})=-\frac{A_{10}\ \ e^{-A_{7} \sqrt{A_{4}}\ +A_4\ t}}{\sqrt{A_{5}}(\sqrt{A_4}+A_8)},$$ $$\gamma_3: -R \rightarrow -\theta_c,\ s=r e^{i\pi},\ s+\theta_c=r e^{i\pi}+\theta_c;\quad \gamma_5: -\theta_c \rightarrow 0,\ s=r e^{i\pi},\ s+\theta_c=r e^{i\pi}+\theta_c; \quad \gamma_7: 0 \rightarrow -\theta_c, \ s=r e^{-i\pi},\ s+\theta_c=r e^{-i\pi}+\theta_c;\quad \gamma_9: -\theta_c \rightarrow -R,\ s=r e^{-i\pi},\ s+\theta_c=(r e^{i\pi}+\theta_c)e^{-2i\pi}.$$ $$\frac{1}{2\pi i}\int_{\gamma_3} F_{5}(s) e^{st}ds = -\frac{A_{10}}{2\pi i}\int_{R}^{\theta_c}\frac{e^{-A_7\sqrt{r}\ e^{i\pi/2}}\ e^{r e^{i\pi t}} e^{i\pi}dr}{(\sqrt{r }e^{i\pi /2}+A_8)\sqrt{r e^{i\pi}+\theta_c}(r e^{i\pi}-A_4)}=-\frac{A_{10}}{2\pi }\int^{R}_{\theta_c}\frac{e^{-iA_7\sqrt{r}}e^{-rt}dr}{(i\sqrt{r}+A_8)\sqrt{r-\theta_c}(r+A_4)},$$

$$\frac{1}{2\pi i}\int_{\gamma_9} F_{5}(s) e^{st}ds = -\frac{A_{10}}{2\pi i}\int^{R}_{\theta_c}\frac{e^{-A_7\sqrt{r}\ e^{-i\pi/2}}\ e^{r e^{-i\pi t}} e^{-i\pi}dr}{(\sqrt{r }e^{-i\pi /2}+A_8)\sqrt{(r e^{i\pi}+\theta_c)e^{-2i\pi}}(r e^{-i\pi}-A_4)}=-\frac{A_{10}}{2\pi }\int^{R}_{\theta_c}\frac{e^{iA_7\sqrt{r}}e^{-rt}dr}{(-i\sqrt{r}+A_8)\sqrt{r-\theta_c}(r+A_4)},$$

$$\frac{1}{2\pi i}\int_{\gamma_5} F_{5}(s) e^{st}ds = -\frac{A_{10}}{2\pi i}\int^{0}_{\theta_c}\frac{e^{-A_7\sqrt{r}e^{i\pi/2}}\ e^{r e^{i\pi t}}\ e^{i\pi}dr}{(\sqrt{r }e^{i\pi /2}+A_8)\sqrt{r e^{i\pi}+\theta_c}(r e^{i\pi}-A_4)}=\frac{A_{10}}{2\pi i}\int_{0}^{\theta_c}\frac{e^{-iA_7\sqrt{r}}e^{-rt}dr}{(i\sqrt{r}+A_8)\sqrt{\theta_c-r}(r+A_4)},$$

$$\frac{1}{2\pi i}\int_{\gamma_7} F_{7}(s) e^{st}ds = -\frac{A_{10}}{2\pi i}\int_{0}^{\theta_c}\frac{e^{-A_7\sqrt{r}\ e^{-i\pi/2}}\ e^{r e^{-i\pi t}} e^{-i\pi}dr}{(\sqrt{r}e^{-i\pi /2}+A_8)\sqrt{r e^{i\pi}+\theta_c}(r e^{-i\pi}-A_4)}=-\frac{A_{10}}{2\pi i}\int_{0}^{\theta_c}\frac{e^{i A_7\sqrt{r}}e^{-rt}dr}{(-i\sqrt{r}+A_8)\sqrt{\theta_c-r}(r+A_4)},$$

$$ILT(F_{5}(s))=-\frac{A_{10}\ \ e^{-A_{7} \sqrt{A_{4}}\ +A_4\ t}}{\sqrt{A_{5}}(\sqrt{A_4}+A_8)} + \frac{A_{10}}{\pi }\int^{R}_{\theta_c}\frac{(A_8 \cos{A_7\sqrt r}-\sqrt r\sin{A_7\sqrt r})e^{-rt}dr}{(r+A_8^2)\sqrt{r-\theta_c}(r+A_4)} + \frac{A_{10}}{\pi }\int_{0}^{\theta_c}\frac{(A_8 \sin{A_7\sqrt r}+\sqrt r\cos{A_7\sqrt r})e^{-rt}dr}{(\sqrt{r}+A_8^2)\sqrt{\theta_c-r}(r+A_4)}.$$

I am not sure of the choice of paths and their parametric expressions.