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TuringTest

  • 3 years ago

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

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  1. TuringTest
    • 3 years ago
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    I've had this one on my head for a couple days now, so whoever can help me gets mad respect :)

  2. experimentX
    • 3 years ago
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    you can do that using partial fraction. but this is not quite interesting.

  3. TuringTest
    • 3 years ago
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    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.

  4. TuringTest
    • 3 years ago
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    Is PF really the only way I guess is my real question

  5. TuringTest
    • 3 years ago
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    it seems like there should be a trick

  6. experimentX
    • 3 years ago
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    yeah ... very short and elegant method.

  7. TuringTest
    • 3 years ago
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    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

  8. TuringTest
    • 3 years ago
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    okay now I see they are the same, so either that was a dumb question, or there is a trick I want to know

  9. experimentX
    • 3 years ago
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    there might be.

  10. experimentX
    • 3 years ago
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    trig subs might work. but this seems more nasty.

  11. dape
    • 3 years ago
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    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.

  12. dape
    • 3 years ago
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    Muhahaha

  13. amistre64
    • 3 years ago
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    ... id have to brute away at it :/

  14. dape
    • 3 years ago
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    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

  15. dape
    • 3 years ago
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    I think it's a more elegant solution, despite it involving some pretty high level stuff.

  16. dape
    • 3 years ago
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    That was fun, when I first saw that MacLaurin series I got some serious chills down my spine.

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

  18. Algebraic!
    • 3 years ago
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    heh, was gonna say, you lost an x!

  19. Algebraic!
    • 3 years ago
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    then i saw you just forgot it

  20. experimentX
    • 3 years ago
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    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.

  21. Algebraic!
    • 3 years ago
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    there's some extra stuff in there...

  22. experimentX
    • 3 years ago
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    I'm not quite well with zeta functions ... looks impossible from Fourier series.

  23. Algebraic!
    • 3 years ago
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    whatever, if it's actually supposed to be there, I can't figure out why..

  24. dape
    • 3 years ago
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    Maybe I forgot something else, did it all on paper and copied it over.

  25. Algebraic!
    • 3 years ago
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    seems it's just as experimentx did it, what's with the first series and the "(1-"

  26. experimentX
    • 3 years ago
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    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.

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

  28. dape
    • 3 years ago
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    Arent the poles \(-1,\sqrt[3]{-1},-(\sqrt[3]{-1})^2\)?

  29. Dido525
    • 3 years ago
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    Isn't this thing just a trig substitution?

  30. Dido525
    • 3 years ago
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    Never mind I see the x cubed. I can't see how to do this without partial fractions :/ .

  31. TuringTest
    • 3 years ago
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    nice @dape that was definitely more what I was looking for, now this integral is interesting again :)

  32. Mikael
    • 3 years ago
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    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

  33. TuringTest
    • 3 years ago
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    Contour integral are something I certainly learn, but I would then like to see it that way too.

  34. TuringTest
    • 3 years ago
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    certainly need to*

  35. Mikael
    • 3 years ago
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    |dw:1348327205764:dw|

  36. TuringTest
    • 3 years ago
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    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.

  37. experimentX
    • 3 years ago
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    @Mikael how to you plan to evaluate that series via contour integrals? Don't you need need to change the series into integral?

  38. Mikael
    • 3 years ago
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    The question - please READ the question of @TuringTest .

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