inverse laplace transform by using complex integral

Question

given function $$f(s)=\frac{1}{s}\frac{\sqrt{s}-1}{\sqrt{s}+1}$$ and $$\int_{0}^{\infty}{\frac{e^{-xt}}{\sqrt{x}(x+1)}dx=\pi e^t {erfc}(\sqrt{t})}$$

my steps:

contour:$A->B->C->D->E->F->A$ anti-clockwise

$AB$ straight vertical down to up $AB$

$BC$ arc with radius of $R$

$CD$ straight horizontal from $-R$ to $-\epsilon$

$DE$ arc with radius of $-\epsilon$

$EF$ straight horizontal from $-\epsilon$ to $-R$

$FA$ arc with radius of $R$

1. $\int_{BC}=\int_{FA}=0$

2.$CD$:

$s(x)=xe^{\pi i}$ where $x \in[R -> \epsilon]$ and $s'(x)=e^{\pi i}=-1$ and $\sqrt{s}=i\sqrt{x}$

thus $$\int_{CD}=\int_{R}^{\epsilon}{\frac{e^{tx{e^{\pi i}}}}{xe^{\pi i}}{\frac{i\sqrt{x}-1}{i\sqrt{x}+1}}{e^{\pi i}}}dx=-\int_{\epsilon}^{R}{\frac{e^{-xt}}{x}{\frac{i\sqrt{x}-1}{i\sqrt{x}+1}}}dx$$

3.$EF$

$s(x)=xe^{-\pi i}$ where $x \in[\epsilon -> R]$ and $s'(x)=e^{-\pi i}=-1$ and $\sqrt{s}=-i\sqrt{x}$

thus $$\int_{EF}=\int_{\epsilon}^{R}{\frac{e^{tx{e^{-\pi i}}}}{xe^{-\pi i}}{\frac{-i\sqrt{x}-1}{-i\sqrt{x}+1}}{e^{-\pi i}}}dx=-\int_{\epsilon}^{R}{\frac{e^{-xt}}{x}{\frac{i\sqrt{x}+1}{i\sqrt{x}-1}}}dx$$

4.$DE$

$s(\theta)=\epsilon e^{i \theta}$ where $\theta\in[\pi -> -\pi]$ and $s'(\theta)=i\epsilon e^{i \theta}$

thus

$$\int_{\pi}^{-\pi}{\frac{e^{t\epsilon e^{i\theta}}}{\epsilon e^{i\theta}}\frac{\sqrt{\epsilon}{ e^{\frac{i\theta}{2}}-1}}{{\sqrt{\epsilon} e^{\frac{i\theta}{2}}+1}}}{i\epsilon e^{i\theta}}d{\theta}=-i\int_{-\pi}^{\pi}{{e^{t\epsilon e^{i \theta}}}{\frac{{\sqrt{\epsilon}}{e^{i \frac{\theta}{2}}} -1}{{\sqrt{\epsilon}}{e^{i \frac{\theta}{2}}} +1}}}d{\theta}$$

over all

$\int_{AB}=-\lim_{R->\infty,\epsilon->0}{[\int_{CD}+\int_{ED}+\int_{EF}]}$

$$\int_{AB}=-\lim_{R->\infty,\epsilon->0}{[-\int_{\epsilon}^{R}{\frac{e^{-xt}}{x}{(\frac{i\sqrt{x}-1}{i\sqrt{x}+1} + \frac{i\sqrt{x}+1}{i\sqrt{x}-1})}}dx - -i\int_{-\pi}^{\pi}{{e^{t\epsilon e^{i \theta}}}{\frac{{\sqrt{\epsilon}}{e^{i \frac{\theta}{2}}} -1}{{\sqrt{\epsilon}}{e^{i \frac{\theta}{2}}} +1}}}d{\theta}}]$$

$$\int_{AB}=\int_{0}^{\infty}{\frac{e^{-xt}}{x}{\frac{2(x-1)}{x+1}}}dx + i\int_{-\pi}^{\pi}{(-1)d{\theta}}$$

$$=2\int_{0}^{\infty}{\frac{e^{-xt}}{x}{({\frac{x}{x+1}}-{\frac{1}{x+1}})}}dx - 2 \pi i$$

$$=2\int_{0}^{\infty}{{({\frac{e^{-xt}}{x+1}}-{\frac{e^{-xt}}{x(x+1)}})}}dx - 2 \pi i$$

$$=2\int_{0}^{\infty}{{{{\sqrt{x}}\frac{e^{-xt}}{\sqrt{x}(x+1)}}-{\frac{1}{\sqrt{x}}\frac{e^{-xt}}{\sqrt{x}(x+1)}}}}dx - 2 \pi i$$

upto here... then .. I cannot do more..

need help...

Answer 1

To answer your question, the most important thing you are leaving out is Cauchy's theorem. You are defining the contour so that it encloses no poles of the LT and is single-valued along itself. Thus, the sum of the contributions along the contours is zero. You should end up with something like

$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} \frac{ds}{s} \frac{\sqrt{s}-1}{\sqrt{s}+1} e^{s t} &= -1 + \frac1{i 2 \pi} \int_0^{\infty} \frac{dx}{x} \left (\frac{i \sqrt{x}-1}{i \sqrt{x}+1} - \frac{-i \sqrt{x}-1}{-i \sqrt{x}+1} \right ) e^{-x t} \\ &= -1 + \frac{2}{\pi} \int_0^{\infty} \frac{dx}{\sqrt{x}} \frac{e^{-x t}}{1+x} \end{align}$$

(I think you lost a sign somewhere.)

I verified the result by Laplace transforming again.