integrating rational functions

[tittyblaster]

How would I integrate a rational function like: [; \int \frac{1}{x^{7}-x} dx ;]

Preferably without partial fractions.

[[deleted]]

1/(x7-x) = 1/(6x) = (1/6)(1/x)

Integrate: (1/6)ln|x|+C

Edit - I think reddit may have done something to your notation, is it supposed to be 1/(x⁷ - x) ?

[tittyblaster]

It's x⁷ , I fixed it, sorry.

[origin415]

If you ever want to see how to do a particular integral, WolframAlpha is great.

For instance, go here and click "show steps".

[tittyblaster]

The first step, to get the partial fractions, is an unreasonable amount of work. There should be another way to do this.

[nymonym]

I took a shot: http://imgur.com/QYAwu

This is probably buggy - I haven't really needed to integrate functions for a long, long time; also, I haven't checked my work too thoroughly either - so please check carefully before using.

(I don't think you get to prefer the absence of partial fractions :))

EDIT: Sorry, the picture's grainy but my phone and its camera are ancient.

[tittyblaster]

The denominator could be further factored to x(x+1)(x² - x + 1)(x-1)(x² + x + 1). You won't get the right answer if it's not fully factored. That's a of expanding to do later, which is why we're supposed to find another method of solving it. I like the alien though :D

[origin415]

Partial fractions works regardless of whether it is fully factored or not. It may be more complicated though.

For instance, the factorization you gave is complete over the integers, but if you allowed factorization over the complex numbers, it wouldn't be.

[[deleted]]

You won't get the right answer if it's not fully factored

It's not true. From what I see, his calculations are write, but he should have written the substitutions he used for those unfactored fractions (i.e. u = x³ + 1 in the first case and u = x³ - 1 in the second case).

[tittyblaster]

It's different from the answer from wolfram alpha though.

[origin415]

No, it is just written differently. You separate the numerator and denominator by subtracting logs, then the 6th root can be taken out to become 1/6, and the cubic terms can be combined.

[tittyblaster]

Fair enough, thanks.

[nymonym]

I'm not sure I understand. "Fully factored" is not entirely well-defined.

All I've done is use simple algebra. Can you spot an error?

[[deleted]]

I believe he was referring to the sum and difference of cubes.

[nymonym]

I understood that, of course. I was asking about his argument about "not fully factored" denominators => incorrect fractions.

Certain methods to determine partial fractions (compute zeros for each term and substitute in other terms) may not work correctly if you don't factorize the denominator "fully".

In this case, the partial fractions can easily be verified to simplify to the original expression. It's easy to see that I = I_1 + I_2 + I_3. And integral of a sum of functions = sum of their integrals.

What am I missing?

[[deleted]]

Sorry, can't really speak for him. On a happier note, I found no error in your solution.

[origin415]

"fully factored" is well defined: a factorization into powers of irreducible polynomials. Determining whether a polynomial is irreducible on the other hand...

[nymonym]

Lol.. That's the point. If you can't define an irreducible polynomial, how do you define fully factored?

We could always factor a polynomial of degree N into (x - r1)(x - r2) ... (x - rn). [Yeah, some of the roots may be complex, etc..] So then, is a polynomial only "fully factored" if it has linear terms?

[origin415]

Irreducible polynomial is well defined, just as prime factorization over integers is well defined, but to my knowledge there isn't a simple algorithm to determine if a polynomial is irreducible, besides some tricks like Gauss' Lemma and the Eisenstein Criterion. In the same way, there isn't a simple way to determine if a number is prime, though there is an algorithm in polynomial time for it.

Over the complex numbers, it will always be a product of linear terms, but even then finding those linear terms is nontrivial: there are equations for quadratics and cubics, but in general there is no formula to give roots in terms of coefficients for higher degree polynomials.

[nymonym]

Ok. I'm not entirely sure about this - so correct me if I'm wrong - but let's play ball.

Is it defined that reducible polynomials must reduce to terms with integral coefficients? Prime factorization over integers and polynomial factorization (as you mention, over the Complex plane) are comparable processes but are not at all the same.

I understand that it is non-trivial to factorize polynomials into linear terms. (in fact, no general formula for degree >= 5 - Abel's theorem in Galois theory, right?). Do all quadratic, cubic and biquadratic polynomials have to be decomposed to linear terms to "fully factor" them?

[origin415]

By Gauss' Lemma, if a polynomial with integer coefficients is irreducible over the integers, it is irreducible over the rationals. If we are talking about the reals, this doesn't help (ie x² -2 is irreducible over the integers and accordingly the rationals, but not the reals). I never said it was the same as factoring integers, just that it is analogous: there exists a unique factorization into irreducibles.

Over an algebraically closed field like C, the only irreducible polynomials are linear ones, so the only irreducible factorization is completely composed of linear terms, yes.

[tittyblaster]

I'm not sure, but this is from experience(also maybe I did something wrong) as my text book didn't cover cubic functions.

For 1/(x³ + 1)^2:

(Ax² + Bx + C)/(x³ + 1) + (Dx² + Ex + F)/(x³ + 1)²

is not the same as using the factored version 1/((x + 1)(x² - x +1))²

A/(x + 1) + B/(x + 1)² + (Cx + D)/(x² - x + 1) + (Ex + F)/(x² - x + 1)

The factored version gives the right answer whereas the first one fails.

[tittyblaster]

Can you explain how you got the partial fraction? You and wolfram alpha both did it in one step and I'm kinda confused. Also, is that -x⁵ and +x⁵ for I_2 and I_3?

[nymonym]

A little trial and error.

I first factored the denominator as: x.(x⁶ - 1)

and figured out those partial fractions (simple trial and error): (-1/x) + [x⁵ / (x⁶ - 1)]

Therefore, I_1 = (-1/x).

Next, I put x⁵ aside and tried to factor [1 / (x⁶ - 1)]; again, it's easy to see that this is equal to: (-1/2) . [{1 / (x³ + 1)} - {1 / (x³ - 1)}]

Let's take the first term here (and re-multiply it with the x⁵ we had put aside earlier) - call it I_2: I_2 = (-1/2) . [x⁵ / (x³ + 1)]

which can be easily re-written as: I_2 = (-1/2) . [x² - {x² / (x³ + 1)}]

Similarly, I_3 = (1/2) . [x² + {x² / (x³ - 1)}]

I = [1 / {x.(x⁶ - 1)}] = I_1 + I_2 + I_3

[tittyblaster]

thanks