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MathGroup Archive 2006

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Re: FoourierTransform of a function defined in sections

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70234] Re: FoourierTransform of a function defined in sections
  • From: dimmechan at yahoo.com
  • Date: Sun, 8 Oct 2006 02:04:57 -0400 (EDT)
  • References: <eg4soa$ffu$1@smc.vnet.net><eg82ch$n2c$1@smc.vnet.net>

In my previous post I forgot the definition of hh[x].

hh[x_] := Piecewise[{{Sin[x]^2, Abs[x] <= Pi}, {0, x > Pi || x < Pi}}]

So

FourierTransform[hh[x], x, s]
-((2*Sqrt[2/Pi]*Sin[Pi*s])/(-4*s + s^3))

(1/Sqrt[2*Pi])*Integrate[hh[x]*Exp[I*s*x], {x, -Infinity, Infinity}]
-((2*Sqrt[2/Pi]*Sin[Pi*s])/(s*(-4 + s^2)))

N[%] /. s -> 1/2
0.8510768648563898

Chop[(1/Sqrt[2*Pi])*NIntegrate[hh[x]*Exp[I*(1/2)*x], {x, -Infinity,
Infinity}]]
0.8510768648563899


Î?/Î? dimmechan at yahoo.com έγÏ?αÏ?ε:
> You could work as follows:
>
> $VersionNumber
> 5.2
>
> FourierTransform[hh[x], x, s]
> -((2*Sqrt[2/Pi]*Sin[Pi*s])/(-4*s + s^3))
>
> or
>
> (1/Sqrt[2*Pi])*Integrate[hh[x]*Exp[I*s*x], {x, -Infinity, Infinity}]
> -((2*Sqrt[2/Pi]*Sin[Pi*s])/(s*(-4 + s^2)))
>
> In either case
>
> N[%] /. s -> 1/2
> 0.8510768648563898
>
> (*check*)
>
> Chop[(1/Sqrt[2*Pi])*NIntegrate[hh[x]*Exp[I*(1/2)*x], {x, -Infinity,
> Infinity}]]
> 0.8510768648563899
> 
> 
> Cheers


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