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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0you can do that using partial fraction. but this is not quite interesting.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0the solution from PF I got seemed drastically different than wolf's elegant answer, though I could have just been unable to manipulate the algebra.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0Is PF really the only way I guess is my real question

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0it seems like there should be a trick

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0yeah ... very short and elegant method.

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0trig subs might work. but this seems more nasty.

dape
 2 years ago
Best ResponseYou've already chosen the best response.3Yeah, 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.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0... id have to brute away at it :/

dape
 2 years ago
Best ResponseYou've already chosen the best response.3It 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

dape
 2 years ago
Best ResponseYou've already chosen the best response.3I think it's a more elegant solution, despite it involving some pretty high level stuff.

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

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

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0this 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.

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not quite well with zeta functions ... looks impossible from Fourier series.

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

dape
 2 years ago
Best ResponseYou've already chosen the best response.3Maybe I forgot something else, did it all on paper and copied it over.

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0still ... 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.

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

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

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0Isn't this thing just a trig substitution?

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0Never mind I see the x cubed. I can't see how to do this without partial fractions :/ .

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

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.0Yes @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
 2 years ago
Best ResponseYou've already chosen the best response.0Contour integral are something I certainly learn, but I would then like to see it that way too.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0Having 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
 2 years ago
Best 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?

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.0The question  please READ the question of @TuringTest .
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