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MathGroup Archive 2006

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Re: does anybody know how to find the inverse Laplace transform of this wierd thing?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64116] Re: [mg64100] does anybody know how to find the inverse Laplace transform of this wierd thing?
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Thu, 2 Feb 2006 00:05:03 -0500 (EST)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

H[s_]=1/s^2*Exp[s^2*a^2/2]*
    Integrate[Exp[-u^2/2], {u , s*a , Infinity}]

(E^((a^2*s^2)/2)*Sqrt[Pi/2]*Erfc[(a*s)/Sqrt[2]])/s^2

h[t_]=InverseLaplaceTransform[H[s], s, t]//
    Simplify

a*(-1 + E^(-(t^2/(2*a^2)))) + Sqrt[Pi/2]*t*
   Erf[t/(Sqrt[2]*a)]

H[s]==FullSimplify[
    LaplaceTransform[h[t], t, s], a>0]

True


Bob Hanlon

> 
> From: "gino" <loseminds at hotmail.com>
To: mathgroup at smc.vnet.net
> Subject: [mg64116] [mg64100]  does anybody know how to find the inverse Laplace 
transform of this wierd thing?
> 
> Want to find the inverse Laplace transform of the following term:
> 
> H(s)=1/s^2*exp(s^2*a^2/2)*integrate(exp(-u^2/2), u from s*a to +infinity)
> 
> How to do that?
> 
> ------------------------------
> 
> Making relaxation to the problem, if I have to find only certain sampled 
> values of the inverse Laplace transform of H(s), let's denote it as h(t),
> 
> I just need to find h(1), h(2), h(3), etc.
> 
> Is there a short cut for it?
> 
> Thanks a lot! 
> 
> 
> 


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