Bug in Mathematica 6 - Integrate - 16 - (Log, Sqrt, false divergence)

Discussion in 'Recreational Math' started by Vladimir Bondarenko, Jun 23, 2007.

  1. Hello again from the VM machine...

    Integrate[Log[z+Sqrt[z^2-1]]/(1+z^2)^3, {z,1,Infinity}]
    NIntegrate[Log[z+Sqrt[z^2-1]]/(1+z^2)^3, {z,1,Infinity}]

    Integral does not converge
    0.0332316


    Best wishes,

    Vladimir Bondarenko

    VM and GEMM architect
    Co-founder, CEO, Mathematical Director

    http://www.cybertester.com/ Cyber Tester, LLC
    http://maple.bug-list.org/ Maple Bugs Encyclopaedia
    http://www.CAS-testing.org/ CAS Testing
     
    Vladimir Bondarenko, Jun 23, 2007
    #1
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  2. Vladimir Bondarenko

    dimitris Guest

    Good morning!

    The integral is not a trivial one.

    Version->5.2

    In[2]:=
    TimeConstrained[Integrate[Log[z + Sqrt[z^2 - 1]]/(1 + z^2)^3, {z, 1,
    Infinity}], 300]
    Out[2]=
    $Aborted

    Note the complexity of the indefinite integral

    In[5]:=
    LeafCount[Integrate[Log[z + Sqrt[z^2 - 1]]/(1 + z^2)^3, z]]
    Out[5]=
    8749

    Dimitris


    Vladimir Bondarenko :
     
    dimitris, Jun 23, 2007
    #2
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  3. Vladimir Bondarenko

    Axel Vogt Guest

    It is (-7/4*2^(1/2)+3*ln(2^(1/2)+1))*Pi/16. Without cheating: -7/64*2^(1/2)*Pi-
    3/16*I*dilog((2^(1/2)+1+I)/(2^(1/2)+1))+3/16*I*dilog((2^(1/2)+1-I)/(2^(1/2)+1))+
    3/16*I*dilog((2^(1/2)-1+I)/(2^(1/2)-1))-3/16*I*dilog((2^(1/2)-1-I)/(2^(1/2)-1))
    where the dilog (of conjugated arguments) have to be simplified (manually?).
     
    Axel Vogt, Jun 23, 2007
    #3
  4. AV> (3*ln(sqrt(2)+1)-7/4*sqrt(2))*Pi/16

    Purr... purrrrr! Like the most delicious
    elite cat's sour cream...

    Author, write even more!!

    (contours? how on galaxies? :)
     
    Vladimir Bondarenko, Jun 23, 2007
    #4
  5. Vladimir Bondarenko

    dimitris Guest

    In version 5.2 after almost 10 minutes in my machine
    I got

    In[1]:=
    Integrate[Log[z + Sqrt[z^2 - 1]]/(1 + z^2)^3, {z, 1, Infinity}]

    Out[1]=
    Integrate[Log[z + Sqrt[-1 + z^2]]/(1 + z^2)^3, {z, 1, Infinity}]

    that is the integral unevaluated.

    Dimitris

    Vladimir Bondarenko :
     
    dimitris, Jun 23, 2007
    #5
  6. Vladimir Bondarenko

    dimitris Guest

    Application of the NL formula fails also
    (at least in Mma 5.2).

    Dimitris

    Vladimir Bondarenko :
     
    dimitris, Jun 23, 2007
    #6
  7. Vladimir Bondarenko

    dimitris Guest

    Hi Axel.

    Could you show us the steps in order to derive the result?

    Thank you very much.

    Greetings from Greece.

    Dimitris

    Axel Vogt :
     
    dimitris, Jun 23, 2007
    #7
  8. Vladimir Bondarenko

    Axel Vogt Guest

    My second identity is being a shaman, so I gave me some Siberian
    mushrooms and during the seance I met dolphins from Kazachya Bay,
    Ukraine. After promising to send purring cats for water polo they
    beamed me into a wood - seemed to be deeply in Canada. I realized
    a huge bear, mildly astonished, he looked over my *left* shoulder
    and grumbled 'Marichev - do you know that guy?' After a 'njet' he
    hit me on my nose. Darkness. I awoke with terrible headache (the
    bear, certainly), but with a small slip of paper in my *right*
    fist and the following message "Da da one: ... (the solution) ...
    And do not forget to send us Vladimir. The Dolphins."
     
    Axel Vogt, Jun 23, 2007
    #8
  9. Vladimir Bondarenko

    Axel Vogt Guest

    It is done in M11 by brute change of variables and simplification:

    Int(ln(z+sqrt(z^2-1))/(1+z^2)^3,z = 1 .. infinity);
    changevar(ln(z+(z^2-1)^(1/2))=x,%,x);
    simplify(%);
    value(%); J:=simplify(%);

    To identify the dilog part multiply by 16/3, evaluate numerical
    and use Plouffe's inverter (had no luck and patients to do it
    the 'exact' way trying dilog ladders).


    PS: please do not top-post in newsgroups.
     
    Axel Vogt, Jun 23, 2007
    #9
  10. Dear Herr Shaman,

    I envy you very much (with *white* envy!)
    Where could I pick up 2-3 big sacks of
    those divine mushrooms?!
     
    Vladimir Bondarenko, Jun 23, 2007
    #10
  11. Vladimir Bondarenko

    Axel Vogt Guest

    :)

     
    Axel Vogt, Jun 23, 2007
    #11
  12. Vladimir Bondarenko

    dimitris Guest

    Dear Alex,

    Thank you very much for showing the amazing solution!
    But what do you mean by
    Regards
    Dimitris

    Axel Vogt :
     
    dimitris, Jun 23, 2007
    #12
  13. Vladimir Bondarenko

    dimitris Guest

    Dear Alex,

    Thank you very much for showing the amazing solution!
    But what do you mean by
    Regards
    Dimitris

    Axel Vogt :
     
    dimitris, Jun 23, 2007
    #13
  14. Vladimir Bondarenko

    Axel Vogt Guest

    In business it is common to reply above the old mails,
    in anti-chronological order (and mostly older mails are
    ignored).

    In 'science' it is common to use the natural ordering
    since humans read sequential (and from top to bottom
    in our culture) and have previous information at hand,
    deleting repetitions if needed.

    That's the convention also used in usenet (VB's makes
    his own threads unreadable, but if he insists ...).

    And for subjects: if they all start with the same long
    text they might be ignored by many (newsreaders often
    are formatted to show only the first 20-30 characters).

    Axel
     
    Axel Vogt, Jun 23, 2007
    #14
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