       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>
• 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*E^(I*m*t - (Gamma*t*Sign[t])/Sqrt)*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.
>