Yatir Halevi

Posted on Monday, 06 January, 2003 - 04:02 pm:

\int 1 / (x^{6} + 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

\pm i

\pm \sqrt{3} / 2 \pm 1 / 2

, which lead to the factorisation of

1 + x^{6}

(1 + x^{2}) (1 - \sqrt{3} x + x^{2}) (1 + \sqrt{3} x + x^{2})

, or

(1 + x^{2}) (1 - x^{2} + x^{4})

(which is obvious, in fact, from the factorisation of

1 + x^{3}

! :) ). You should hopefully now be able to decompose

1 / (1 + x^{6})

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:

$\int 1 / (x^{6} + 1) = \tan^{- 1} x + 1 / 2 \tan^{- 1} [x / (1 - x^{2})] + \sqrt{3} / 4 \log | (1 + \sqrt{3} x + x^{2}) / (1 - \sqrt{3} x + x^{2}) |$
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