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! > > >