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MathGroup Archive 2005

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Re: "teach" mathematica an integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54382] Re: "teach" mathematica an integral
  • From: "Scout" <user at domain.com>
  • Date: Sat, 19 Feb 2005 02:32:55 -0500 (EST)
  • References: <curp29$qvk$1@smc.vnet.net> <cv08ns$j7e$1@smc.vnet.net> <cv2dgp$2dp$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

> LaplaceTransform[Erf[(a - 2*b*x)/(2*Sqrt[c*x])], x, s]

Have you tried with this Laplace property?

L[f '(t)] = s * L[f(t)] - f(0)

In[1]:=
\!\(D[Erf[\(a - 2\ b\ x\)\/\(2\ \@\(c\ x\)\)], x]\)

Out[1]=
\!\(\(2\ \[ExponentialE]\^\(-\(\((a - 2\ b\ x)\)\^2\/\(4\ c\ x\)\)\)\ \
\((\(-\(b\/\@\(c\ x\)\)\) - \(c\ \((a - 2\ b\ x)\)\)\/\(4\ \((c\ \
x)\)\^\(3/2\)\))\)\)\/\@\[Pi]\)

In[2]:=
LaplaceTransform[%1, x, s]

Out[2]=
\!\(\(2\ \((\(-\(\(a\ \[ExponentialE]\^\(\(a\ b\)\/c - \@\(a\^2\/c\)\ \
\@\(b\^2\/c + s\)\)\ \@\[Pi]\)\/\(2\ \@\(a\^2\/c\)\ \@c\)\)\) - \(b\ \
\[ExponentialE]\^\(\(a\ b\)\/c - \@\(a\^2\/c\)\ \@\(b\^2\/c + s\)\)\ \
\@\[Pi]\)\/\(2\ \@c\ \@\(b\^2\/c + s\)\))\)\)\/\@\[Pi]\)

In[3]:=
Simplify[(%2 + Erf[0])/s]

Out[3]=
\!\(\(-\(\(\[ExponentialE]\^\(\(a\ b\)\/c - \@\(a\^2\/c\)\ \@\(b\^2\/c + \
s\)\)\ \((b\ \@\(a\^2\/c\) +
              a\ \@\(b\^2\/c + s\))\)\)\/\(\@\(a\^2\/c\)\ \@c\ s\ 
\@\(b\^2\/c \
+ s\)\)\)\)\)

Does It help you?

~Scout~

"julia" <dbug at hotmail.de> ha scritto nel messaggio 
news:cv2dgp$2dp$1 at smc.vnet.net...
> Hi,
>
> The term of a model, i need to transform is
> (0.25*a^2*Erf[(a - 2*b*x)/(2*(c*x)^0.5)])/(b^2*x^3)
>
> (a,b,c  are parameters)
>
> The transformation, which mathematica didn't do was
>
> LaplaceTransform[Erf[(a - 2*b*x)/(2*Sqrt[c*x])], x, s]
>
> I performed the transformation by partial integration. This leads to
>
> ((-b)*E^((a*(b - a*Sqrt[c/a^2]*Sqrt[b^2/c + s]))/c) + (Sqrt[c] -
> a*Sqrt[c/a^2]*E^((a*(b - a*Sqrt[c/a^2]*Sqrt[b^2/c + s]))/c))*
> Sqrt[b^2/c + s])/(Sqrt[c]*s*Sqrt[b^2/c + s])
>
> now i want to transform the first term. The Problem is that the
> Erf-Expression is part of a product...
>
>
>


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