Inverse Laplace Transform as Bromwich Integral-6

$$F(s)=\frac{\sqrt{s+\alpha }}{s+\beta }(\alpha >\beta >0)$$

일 때 inverse Laplace transform은 Bromwich integral을 써서 표현하면 $$f(t)=\frac{1}{2\pi i}{\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}\frac{\sqrt{s+\alpha }}{s+\beta }{e}^{st}ds$$

$F(s)$가 $s=-\beta$에서 simple pole을 가지며 $s=-\alpha$은 branch point에 해당한다. 이 적분을 수행하기 위해서 그림과 같은 경로를 잡자.

Branch cut은 $s=-\alpha$에서 시작하여 $-x$축 방향으로 선택한다. 그러면 $$\frac{1}{2\pi i}\oint F(s)e^{st}ds=\frac{1}{2\pi i}\times 2\pi i\times \text{Res}(s=-\beta )$$이다. $s=-\beta$에서 residue는 

$$\text{Res}(s=-\beta )=\sqrt{\alpha -\beta }{e}^{-beta}$$이다. Branch cut을 감싸는 경로 $C_1,C_2$에서 적분을 수행하기 위해서 $z=s+\alpha$, $\gamma \equiv \alpha -\beta$로 놓으면

$$\frac{1}{2\pi i}{\int }_{C_1+C_2}=\frac{1}{2\pi i}{\int }_{C_1}\frac{\sqrt{z}}{z-\gamma }{e}^{(z-\alpha )t}dz+\frac{1}{2\pi i}{\int }_{C_2}\frac{\sqrt{z}}{z-\gamma }{e}^{(z-\alpha )t}dz$$

$C_1$에서 $z=u{e}^{i\pi }(u:\mathrm{\infty }\to 0)$, $C_2$에서는 $z=u{e}^{-i\pi }(u:0\to \mathrm{\infty })$이므로 

$$\begin{array}{c}\frac{1}{2\pi i}{\int }_{C_1+C_2}=\frac{1}{2\pi i}[{\int }_{\mathrm{\infty }}^0\frac{\sqrt{u}{e}^{i\pi /2}{e}^{-ut}(-du)}{-u-\gamma }+{\int }_0^{\mathrm{\infty }}\frac{\sqrt{u}{e}^{-i\pi /2}{e}^{-ut}(-du)}{-u-\gamma }]\\ =-\frac{e^{-\gamma t}}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\sqrt{u}{e}^{-ut}du}{u+\gamma }\end{array}$$

$$\begin{array}{c}\to f(t)=\text{Res}(-\beta )-\frac{1}{2\pi i}{\int }_{C_1+C_2}\\ =\sqrt{\alpha -\beta }{e}^{-\beta t}+\frac{1}{\pi }{e}^{-\alpha t}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{u}}{u+\gamma }{e}^{-ut}du\\ =\sqrt{\alpha -\beta }{e}^{-\beta t}+\frac{\sqrt{\gamma }}{\pi }{e}^{-\alpha t}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{1+x}{e}^{-(\gamma t)x}dx(\leftarrow u\equiv \gamma x)\end{array}$$

그런데 $\frac{\sqrt{x}}{1+x}=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}(1+x)}$임을 이용하면$$\begin{gathered} A\equiv {\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{1+x}{e}^{-(\gamma t)x}dx \\ ={\int }_0^{\mathrm{\infty }}\frac{e^{-\gamma tx}dx}{\sqrt{x}}-{\int }_0^{\mathrm{\infty }}\frac{e^{-\gamma tx}dx}{\sqrt{x}(1+x)}=\sqrt{\frac{\pi }{\gamma t}}-B \end{gathered}$$ $$\begin{gathered} B\equiv {\int }_0^{\mathrm{\infty }}\frac{e^{-\gamma tx}dx}{\sqrt{x}(x+1)}=e^{\gamma t}{\int }_0^{\mathrm{\infty }}\frac{e^{-(1+x)\gamma t}dx}{\sqrt{x}(1+x)} \\ =e^{\gamma t}\times \pi \times \text{erfc}(\sqrt{\gamma t}) \end{gathered}$$ $$\to A=\sqrt{\frac{\pi }{\gamma t}}[1-\sqrt{\pi \gamma t}{e}^{\gamma t}\text{erfc}(\sqrt{\gamma t})]$$따라서 $F(s)$의 inverse Laplace transform은

$$\begin{array}{rl}f(t)& =\frac{1}{2\pi i}{\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}\frac{\sqrt{s+\alpha }}{s+\beta }{e}^{st}ds\\ & =\sqrt{\alpha -\beta }{e}^{-\beta t}+\frac{e^{-\alpha t}}{\sqrt{\pi t}}[1-\sqrt{\pi (\alpha -\beta )t}{e}^{(\alpha -\beta )t}\text{erfc}\left(\sqrt{(\alpha -\beta )t}\right)]\end{array}$$