6
$\begingroup$

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...

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ got it..thanks so much.... that's a serious mistake... $\endgroup$
    – leave2014
    Oct 20, 2014 at 15:01
  • $\begingroup$ I lost a "-" at the EF term..$$\int_{EF}=-\int_{\epsilon}^{R}{\frac{e^{-xt}}{x}{\frac{-i\sqrt{x}-1}{i \sqrt{x}-1}}}dx$$ $\endgroup$
    – leave2014
    Oct 20, 2014 at 15:03
  • $\begingroup$ @leave2014 : how did you solve the last integral $$\int_0^{\infty} \frac{dx}{\sqrt{x}} \frac{e^{-x t}}{1+x}$$ $\endgroup$
    – Wita
    Apr 17, 2015 at 8:30
  • $\begingroup$ @RonGordon: maybe you know how the last integral can be solved .. I do not know which substitution I should use when the integral has $x^{3/2}$ term $\endgroup$
    – Wita
    Apr 17, 2015 at 8:34
  • $\begingroup$ @2che: to evaluate the integral, I subbed $x=u^2$ to get $$\int_{-\infty}^{\infty} du \frac{e^{-t u^2}}{1+u^2} $$ which may be evaluated using Fourier transform techniques or by a differentiation trick. $\endgroup$
    – Ron Gordon
    Apr 17, 2015 at 9:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .