       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)
• 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])/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*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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