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Re: "teach" mathematica an integral

• To: mathgroup at smc.vnet.net
• Subject: [mg54635] Re: "teach" mathematica an integral
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Thu, 24 Feb 2005 03:21:45 -0500 (EST)
• Organization: The University of Western Australia
• References: <curp29$qvk$1@smc.vnet.net> <cv08ns$j7e$1@smc.vnet.net> <cv2dgp$2dp$1@smc.vnet.net> <cv6suf$6qv$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

In article <cv6suf$6qv$1 at smc.vnet.net>, "Scout" <user at domain.com> 
wrote:

> > 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)

One must take care in evaluating f[0]. For real parameters with a > 0 
and c > 0, as x -> 0

  Erf[(a - 2 b x)/(2 Sqrt[c x])] -> Erf[a/(2 Sqrt[c x])] -> 1

> 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]

This is not correct. You mean Erf[Infinity], which is 1.

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 6488 2734
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