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!
>
>
>