Here's the question you clicked on:
TuringTest
Definite integral\[\int_0^1\frac x{x^3+1}dx\]
I've had this one on my head for a couple days now, so whoever can help me gets mad respect :)
you can do that using partial fraction. but this is not quite interesting.
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.
Is PF really the only way I guess is my real question
it seems like there should be a trick
yeah ... very short and elegant method.
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
okay now I see they are the same, so either that was a dumb question, or there is a trick I want to know
trig subs might work. but this seems more nasty.
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.
... id have to brute away at it :/
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
I think it's a more elegant solution, despite it involving some pretty high level stuff.
That was fun, when I first saw that MacLaurin series I got some serious chills down my spine.
Oops, typo, the integral term on second line should be: \[ -\sum_{n=0}^\infty{\frac{(-1)^n}{3n+1}}\int_0^1{x^{3n+1}\,dx} \]
heh, was gonna say, you lost an x!
then i saw you just forgot it
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.
there's some extra stuff in there...
I'm not quite well with zeta functions ... looks impossible from Fourier series.
whatever, if it's actually supposed to be there, I can't figure out why..
Maybe I forgot something else, did it all on paper and copied it over.
seems it's just as experimentx did it, what's with the first series and the "(1-"
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.
If you've gone McLaurin - then why not go full complex contour integration. The poles are obvious -1, +third-root(-1), -third-root(-1)
Arent the poles \(-1,\sqrt[3]{-1},-(\sqrt[3]{-1})^2\)?
Isn't this thing just a trig substitution?
Never mind I see the x cubed. I can't see how to do this without partial fractions :/ .
nice @dape that was definitely more what I was looking for, now this integral is interesting again :)
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
Contour integral are something I certainly learn, but I would then like to see it that way too.
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.
@Mikael how to you plan to evaluate that series via contour integrals? Don't you need need to change the series into integral?
The question - please READ the question of @TuringTest .