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TuringTestBest ResponseYou've already chosen the best response.0
I've had this one on my head for a couple days now, so whoever can help me gets mad respect :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
you can do that using partial fraction. but this is not quite interesting.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
Is PF really the only way I guess is my real question
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
it seems like there should be a trick
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
yeah ... very short and elegant method.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
okay now I see they are the same, so either that was a dumb question, or there is a trick I want to know
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
trig subs might work. but this seems more nasty.
 one year ago

dapeBest ResponseYou've already chosen the best response.3
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.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
... id have to brute away at it :/
 one year ago

dapeBest ResponseYou've already chosen the best response.3
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: \[ \frac{x}{x^3+1}=xx^4+x^7x^{10}+...=x\sum_{n=0}^\infty(1)^nx^{3n} \] 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 zetafunctions in combination with some subtitutions, but at least it comes out the same as wolf's answer :p
 one year ago

dapeBest ResponseYou've already chosen the best response.3
I think it's a more elegant solution, despite it involving some pretty high level stuff.
 one year ago

dapeBest ResponseYou've already chosen the best response.3
That was fun, when I first saw that MacLaurin series I got some serious chills down my spine.
 one year ago

dapeBest ResponseYou've already chosen the best response.3
Oops, typo, the integral term on second line should be: \[ \sum_{n=0}^\infty{\frac{(1)^n}{3n+1}}\int_0^1{x^{3n+1}\,dx} \]
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
heh, was gonna say, you lost an x!
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
then i saw you just forgot it
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
there's some extra stuff in there...
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
I'm not quite well with zeta functions ... looks impossible from Fourier series.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
whatever, if it's actually supposed to be there, I can't figure out why..
 one year ago

dapeBest ResponseYou've already chosen the best response.3
Maybe I forgot something else, did it all on paper and copied it over.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
seems it's just as experimentx did it, what's with the first series and the "(1"
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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.
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
If you've gone McLaurin  then why not go full complex contour integration. The poles are obvious 1, +thirdroot(1), thirdroot(1)
 one year ago

dapeBest ResponseYou've already chosen the best response.3
Arent the poles \(1,\sqrt[3]{1},(\sqrt[3]{1})^2\)?
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Isn't this thing just a trig substitution?
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Never mind I see the x cubed. I can't see how to do this without partial fractions :/ .
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
nice @dape that was definitely more what I was looking for, now this integral is interesting again :)
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
Contour integral are something I certainly learn, but I would then like to see it that way too.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
@Mikael how to you plan to evaluate that series via contour integrals? Don't you need need to change the series into integral?
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
The question  please READ the question of @TuringTest .
 one year ago
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