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 > > >