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What is the Laplace transform of this function? $$f(x)=\frac{\sqrt{x}}{1+x}$$

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Maybe just go from the definition? I don't really see any clever way to do it. How far have you gotten? Have you tried the substitution $\sqrt x = t$ to see where that goes? –  Scaramouche Jan 14 '12 at 8:47
    
According to the Wolfram Alpha result is : $\sqrt{\frac{\pi}{s}}-\pi \cdot e^s \cdot erfc(\sqrt{s})$ , where erfc is complementary error function... –  pedja Jan 14 '12 at 8:55

2 Answers 2

up vote 6 down vote accepted

We want $\displaystyle F(t)=\int^{\infty}_0 \frac{\sqrt{x}}{1+x} e^{-t x} dx$

$F(t) e^{-t}=\int^{\infty}_0 \frac{\sqrt{x}}{1+x} e^{-t(x+1)} dx$
$\frac{d}{dt}(F(t) e^{-t})=-\int^{\infty}_0 \sqrt{x}e^{-t(x+1)} dx$
$e^t\frac{d}{dt}(F(t) e^{-t})=-\int^{\infty}_0 \sqrt{x}e^{-tx} dx=-t^{-1/2-1}\Gamma(1/2+1)=-\sqrt{\pi}/(2t^{\frac32})$
because you got a specific case of the integral defining the Gamma function.

After that you'll have to revert the operations :
$F(t)= -\frac{\sqrt{\pi}}2 e^t \left(C+\int \frac{e^{-t}}{t\sqrt{t}} dt\right)$
$F(t)= -\frac{\sqrt{\pi}}2 e^t \left(C-\frac{2e^{-t}}{\sqrt{t}}-\int \frac{2e^{-t}}{\sqrt{t}} dt\right)$
$F(t)= -\frac{\sqrt{\pi}}2 e^t \left(C-\frac{2e^{-t}}{\sqrt{t}}-2\int e^{-u^2} du\right)$
where we recognize the integral expression of the Error function multiplied by $\sqrt{\pi}$.

I'll let you reverify all this and determine the constant $C$.

The answer should be $\sqrt{\frac{\pi}{t}} -\pi e^t \text{ erfc}(\sqrt{t})$ for $\Re(t)\gt0$ (with $\text{erfc}(x)=1-\text{erf}(x)$).

(short way : Wolfram Alpha).

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Thank you for your answer! –  robert Jan 14 '12 at 12:01

$$ f(t)=\frac{\sqrt{t}}{1+t} $$ $$ (1+t)f(t)=\sqrt{t} $$ $$ \downarrow {\mathcal{L}} $$ $$ F(s)-\frac{d}{ds}F(s)=\frac{\frac{\sqrt{\pi}}{2}}{s^{\frac{3}{2}}} $$ we have First Order Differential Equation: $$ \frac{d}{ds}F(s)-F(s)=-\frac{\frac{\sqrt{\pi}}{2}}{s^{\frac{3}{2}}} $$ We know : $$ \dfrac{dy}{dt}+ P(t) y = Q(t) \space , \space y=e^{-\int P(t)dt} \left[\int Q(t) e^{\int P(t)dt} dt+c \right] $$ So : $$ F(s)=e^{-\int (-1)ds}\left[ \int \frac{\frac{-\sqrt{\pi}}{2}}{s^{\frac{3}{2}}}e^{\int(-1)ds}ds+C \right] $$ $$ F(s)=e^s \left[ (\frac{-\sqrt{\pi}}{2}) \int \frac{1}{s^{\frac{3}{2}}} e^{-s} ds +C\right] $$ $$ F(s)=e^s \left[ (\frac{-\sqrt{\pi}}{2}) \left( \frac{-2e^{-s}}{\sqrt{s}} -2\sqrt{\pi} .Erf (\sqrt{s}) \right) +C \right]$$ $$ F(s)=\sqrt{\frac{\pi}{s}}+\pi e^s.Erf(\sqrt{s})+Ce^s $$

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+1. Good attempt Amir. ;-) –  Babak S. Dec 22 '12 at 18:13

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