Re: Characteristic function of Caucy distrbution
- To: mathgroup at smc.vnet.net
- Subject: [mg35527] Re: Characteristic function of Caucy distrbution
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Thu, 18 Jul 2002 03:06:08 -0400 (EDT)
- Organization: Universitaet Leipzig
- References: <ah24j0$r3d$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
In[]:=expr = Integrate[
Exp[I t x] Gamma/(Gamma^2/2 + (x - m)^2), {x, -Infinity,
Infinity}]/Pi;
and
In[]:=expr /. a_.*Integrate[ex_, bound_] :>
Integrate @@ {ex /. x -> y + m, bound /. x -> y,
Assumptions -> {Im[t] == 0 && Gamma > 0}} //
FullSimplify[#, Element[Gamma, Reals]] &
Out[]=(Sqrt[2]*E^(I*m*t - (Gamma*t*Sign[t])/Sqrt[2])*Pi)/Gamma
and t*Sign[t] is Abs[t]
Regards
Jens
milkcart wrote:
>
> Hello
>
> How is Characteristic function of Caucy distrbution derived properly.
> I wrote this code as
>
> Simplify[Integrate[E^(I*t*x)/
> (1 + (x - $B&L(B)^2/$B&R(B^2),
> {x, -Infinity,
> Plus[Infinity]}]/
> (Pi*$B&R(B)]
>
> . But this integration return itself.
>
> I know that right answer is Exp[I $B&L(B t -$B&R(B Abs[t]].
> How can I get right answer.
>
> Thanks in advance.
>
> *****************
> milkcart
> milkcart at m17.alpha-net.ne.jp
> *************************