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MathGroup Archive 2005

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Re: Integration under Mathematica 5.0

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57434] Re: [mg57417] Integration under Mathematica 5.0
  • From: "Owen, HL \(Hywel\)" <h.l.owen at dl.ac.uk>
  • Date: Sat, 28 May 2005 05:38:57 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Yes, I get this behaviour in 5.1

Integrate[Sqrt[1 + Sqrt[u]]/(
      E^(a*u)*(2*Sqrt[2]*Sqrt[u])), {u, 0, 1}, Assumptions -> a > 0]

gives

(1/3)*(4 - Sqrt[2])

> -----Original Message-----
> From: José Carlos Santos [mailto:jcsantos at fc.up.pt]
To: mathgroup at smc.vnet.net
> Sent: 27 May 2005 09:57
> Subject: [mg57434] [mg57417] Integration under Mathematica 5.0
> 
> 
> 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
> 
> 
> 


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