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MathGroup Archive 2005

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Re: Integration under Mathematica 5.0

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57453] Re: [mg57417] Integration under Mathematica 5.0
  • From: Murray Eisenberg <murray at math.umass.edu>
  • Date: Sat, 28 May 2005 05:39:24 -0400 (EDT)
  • Organization: Mathematics & Statistics, Univ. of Mass./Amherst
  • References: <200505270856.EAA07672@smc.vnet.net>
  • Reply-to: murray at math.umass.edu
  • Sender: owner-wri-mathgroup at wolfram.com

    $Version
5.1 for Microsoft Windows (January 27, 2005)

   Integrate[Sqrt[1 +
     Sqrt[u]]/(E^(a*u)*(2*Sqrt[2]*Sqrt[u])), {u, 0, 1}] // InputForm
If[Re[a] > 0, (4*(-1 + 2*Sqrt[2]))/3,
   Integrate[Sqrt[1 + Sqrt[u]]/(E^(a*u)*Sqrt[u]),
    {u, 0, 1}, Assumptions -> Re[a] <= 0]]/(2*Sqrt[2])

Happier with that?

José Carlos Santos wrote:
> Hi all,
> 
> At another newsgroup, someone has transcribed this Mathematica
> session:
> 
> In[1]:=$Version
> 
> Out[1]=5.0 for Microsoft Windows (November 18, 2003)
> 
> In[2]:=Integrate[Sqrt[1+Sqrt[u]]/(E^(a*u)*(2*Sqrt[2]*Sqrt[u])),{u,0,1}]
> 
> Out[2]=(1/3)*(4 - Sqrt[2])
> 
> This makes no sense, because the integral does depend upon the parameter
> _a_ (just ask Mathematica to plot the graph). I do not have access to
> Mathematica 5.0, but under the version 4.0 I do not get that answer;
> what I get is the same integral written in a slightly different form.
> 
> Can someone reproduce that behaviour? And, if it is possible to
> reproduce it, can someone *explain* it?
> 
> BTW, the value (1/3)*(4 - Sqrt[2]) is the value of the integral when
> a = 0.
> 
> Best regards,
> 
> Jose Carlos Santos
> 
> 

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305



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