Definite integral\[\int_0^1\frac x{x^3+1}dx\]

- TuringTest

Definite integral\[\int_0^1\frac x{x^3+1}dx\]

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- schrodinger

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- TuringTest

I've had this one on my head for a couple days now, so whoever can help me gets mad respect :)

- experimentX

you can do that using partial fraction. but this is not quite interesting.

- TuringTest

the solution from PF I got seemed drastically different than wolf's elegant answer, though I could have just been unable to manipulate the algebra.

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## More answers

- TuringTest

Is PF really the only way I guess is my real question

- TuringTest

it seems like there should be a trick

- experimentX

yeah ... very short and elegant method.

- TuringTest

I know, that is why I want something besides hacking away with PF which gives
http://www.wolframalpha.com/input/?i=integral+x%2F%281%2Bx%5E3%29

- TuringTest

okay now I see they are the same, so either that was a dumb question, or there is a trick I want to know

- experimentX

there might be.

- experimentX

trig subs might work. but this seems more nasty.

- dape

Yeah, PF would definitely work. Probably won't be able to do much, but I'll try to poke around and see if I can find a nicer method.

- dape

Muhahaha

- amistre64

... id have to brute away at it :/

- dape

It turns out after researching the integrand for a bit that it has an \(extremely\) nice McLaurin series. I don't have the energy to type out the entire derivation, but you can check it's:
\[ \frac{x}{x^3+1}=x-x^4+x^7-x^{10}+...=x\sum_{n=0}^\infty(-1)^nx^{3n} \]
Throwing in this in the integral and doing a partial integration we get:
\[ \int_0^1{\frac{x}{x^3+1}\, dx}=\int_0^1{x\sum_{n=0}^\infty(-1)^nx^{3n}\,dx} \\
=\sum_{n=0}^\infty \frac{(-1)^n}{3n+1}-\sum_{n=0}^\infty(-1)^n\int_0^1{x^{3n}\,dx} \\
=\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}\left(1-\frac{1}{3n+2}\right) \\
=\sum_{n=0}^\infty\frac{(-1)^n}{3n+2} \]
Now that sum is a bit harder to calculate, it involves zeta-functions in combination with some subtitutions, but at least it comes out the same as wolf's answer :p

- dape

I think it's a more elegant solution, despite it involving some pretty high level stuff.

- dape

That was fun, when I first saw that MacLaurin series I got some serious chills down my spine.

- dape

Oops, typo, the integral term on second line should be:
\[ -\sum_{n=0}^\infty{\frac{(-1)^n}{3n+1}}\int_0^1{x^{3n+1}\,dx} \]

- anonymous

heh, was gonna say, you lost an x!

- anonymous

then i saw you just forgot it

- experimentX

this seems pretty straight forward
\[ \int_0^1{x\sum_{n=0}^\infty(-1)^nx^{3n}\,dx} = \sum_{n=0}^\infty(-1)^n \int_0^1x^{3n+1}\,dx = \sum_{n=0}^\infty {(-1)^n \over 3n+2}\]
I get chills when i see these.

- anonymous

there's some extra stuff in there...

- experimentX

I'm not quite well with zeta functions ... looks impossible from Fourier series.

- anonymous

whatever, if it's actually supposed to be there, I can't figure out why..

- dape

Maybe I forgot something else, did it all on paper and copied it over.

- anonymous

seems it's just as experimentx did it, what's with the first series and the "(1-"

- experimentX

still ... i never though of this technique. I haven't been using this technique lately. I'll try to think of this way too in the future. gotta sleep ... nearly 3.30 am here.

- anonymous

If you've gone McLaurin - then why not go full complex contour integration.
The poles are obvious -1, +third-root(-1), -third-root(-1)

- dape

Arent the poles \(-1,\sqrt[3]{-1},-(\sqrt[3]{-1})^2\)?

- anonymous

Isn't this thing just a trig substitution?

- anonymous

Never mind I see the x cubed. I can't see how to do this without partial fractions :/ .

- TuringTest

nice @dape that was definitely more what I was looking for, now this integral is interesting again :)

- anonymous

Yes @dape these are the poles. Yet I am not certain there is an effective choice of contour so that we can find the needed contribution

- TuringTest

Contour integral are something I certainly learn, but I would then like to see it that way too.

- TuringTest

certainly need to*

- anonymous

|dw:1348327205764:dw|

- TuringTest

Having never taken complex analysis I really don't want to pretend that I understand much about integrating in the complex plane, though of course I understand the concept of writing complex numbers in them and roots of unity... basically.

- experimentX

@Mikael how to you plan to evaluate that series via contour integrals? Don't you need need to change the series into integral?

- anonymous

The question - please READ the question of @TuringTest .

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