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MathGroup Archive 2006

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Re: FourierSinTransform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70451] Re: FourierSinTransform
  • From: ab_def at prontomail.com
  • Date: Mon, 16 Oct 2006 02:36:43 -0400 (EDT)
  • References: <egi1sl$jbe$1@smc.vnet.net>

dimmechan at yahoo.com wrote:
> Consider the following function
>
> h[ξ_] := Sqrt[1 + l^2*ξ^2]/ξ
>
> Then (the integral exists in the Hadamard sense)
>
> FourierSinTransform[h[ξ], ξ, x]
> -(MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, x^2/(4*l^2)]/(2*Sqrt[2*Pi]))
>
> Also
>
> FullSimplify[FourierSinTransform[%, x, ξ], {l > 0, ξ > 0}]
> Sqrt[1 + l^2*ξ^2]/ξ
>
> Can I trust the first result?

To evaluate the transform, start with the function h[t] - Sqrt[l^2];
its product with Sin[p*t] can be integrated as an ordinary function.
After the change of variables t = t/Abs[l] we obtain

Sqrt[2/Pi]*Integrate[(Sqrt[1 + t^2]/t - 1)*Sin[p*t/Abs[l]],
  {t, 0, Infinity}] ==

  Sqrt[2/Pi]*Integrate[1/(t*(Sqrt[1 + t^2] + t))*Sin[p*t/Abs[l]],
    {t, 0, Infinity}]

In[1]:= Sqrt[2/Pi]*Integrate[1/(t*(Sqrt[1 + t^2] + t))*Sin[p*t/Abs[l]],
    {t, 0, Infinity}, Assumptions -> p > 0] +
  FourierSinTransform[Sqrt[l^2], t, p]

Out[1]= (Sqrt[l^2]*Sqrt[2/Pi])/p + MeijerG[{{1/2, 1}, {}}, {{1/2, 1/2},
{-(1/2), 0}}, p^2/(4*Abs[l]^2)]/(2*Sqrt[2*Pi])

In[2]:= % /. {l -> -3.3, p -> 2.2}

Out[2]= 1.7449384

In[3]:= -MeijerG[{{1}, {}}, {{-1/2, 1/2}, {0}}, p^2/(4*l^2)]/
  (2*Sqrt[2*Pi]) /. {l -> -3.3, p -> 2.2}

Out[3]= 1.7449384

Maxim Rytin
m.r at inbox.ru


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