# 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,

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

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2. ### dimitrisGuest

Good morning!

The integral is not a trivial one.

Version->5.2

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

Note the complexity of the indefinite integral

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

Dimitris

dimitris, Jun 23, 2007

3. ### Axel VogtGuest

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
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? 5. ### dimitrisGuest

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

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

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

that is the integral unevaluated.

Dimitris

dimitris, Jun 23, 2007
6. ### dimitrisGuest

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

Dimitris

dimitris, Jun 23, 2007
7. ### dimitrisGuest

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
8. ### Axel VogtGuest

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
9. ### Axel VogtGuest

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
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?!

11. ### Axel VogtGuest Axel Vogt, Jun 23, 2007
12. ### dimitrisGuest

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. ### dimitrisGuest

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
14. ### Axel VogtGuest

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