Yatir Halevi
Posted on Monday, 06 January, 2003 - 04:02 pm:
ò1/(x6+1) dx
Marcos
Posted on Monday, 06 January, 2003 - 05:16 pm:
I don't have time to try it myself right now, but this might help:

1/(x6 + 1)
= 1/(x2 + 1)(x4 - x2 + 1)
= 1/3(x2 + 1) + (2 - x2 )/3(x4 - x2 + 1)

Marcos
Paul Smith
Posted on Monday, 06 January, 2003 - 05:25 pm:
The complex 6th roots of -1 are ±i, ±Ö3/2±1/2, which lead to the factorisation of 1+x6 as (1+x2)(1-Ö3x +x2)(1+Ö3x+x2), or (1+x2)(1-x2+x4) (which is obvious, in fact, from the factorisation of 1+x3! :) ). You should hopefully now be able to decompose 1/(1+x6) into partial fractions, each of whose denominators is of degree 2.

Is there a more elegant method, though?

Paul
Marcos
Posted on Monday, 06 January, 2003 - 06:07 pm:
Doing it the long (and horrible/tedious) way I get:

ò1/(x6+1)=tan-1x+1/2 tan-1[x/(1-x2)] +Ö3/4 log|(1+Ö3x+x2)/(1-Ö3x +x2)|
Chances are I've made a few gross slip-ups somewhere in that (I'm not sure I want to try to verify it by differentiation)... I did it, not because I like torturing myself, but because armed with solution (or a poor example of it) maybe someone can spot some sort of elegant method to do the question...

Marcos

Arun Iyer
Posted on Monday, 06 January, 2003 - 06:59 pm:
Marcos,
u just missed the factor of 1/3 (i.e multiply the entire answer by 1/3)... rest is right.

As for an elegant solution,i don't think there is an elegant method to evaluate this indefinite integral.However if this were a definite integral (say with limits -infinity to +infinity or say with limits 0 to infinity) then finding that value is far much easier.

love arun
Marcos
Posted on Monday, 06 January, 2003 - 07:21 pm:
Oops Arun... you're right I just spotted that in my working Thanks
Yatir Halevi
Posted on Monday, 06 January, 2003 - 11:05 pm:
I was thinking along the same lines...although I hoped someone will show me a more elegant solution....
Hauke Worpel
Posted on Friday, 10 January, 2003 - 06:45 am:
There is a very elegant way to do this as long as x is always between 0 and 1. Probably won't apply to you though. Here's how it goes:

1/(1+x6 )=1/(1-u) where u=-x6

Now we use the fact that 1/(1-u)=1+u+u2 +u3 +... to obtain in terms of x

1-x6 +x12 -x18 +...

Integrate term by term to obtain

C+x-6x7 +12x13 -18x19 +...

This gives a very nice series solution, but again it only works if 0 < x < 1.
Arun Iyer
Posted on Friday, 10 January, 2003 - 06:49 am:
if the limits were from 0 to 1,you can convert the integral into a Beta integral as well.

love arun