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How do I evaluate this?

  1. May 24, 2014 #1
    So I was trying to evaluate ∫x(e^x^3) dx
    I know that you use integration by parts but something stumps me.

    I don't know how to integrate e^x^3!!!! I know how to integrate e^3x but not e^x^3!
    Can someone please tell me how to even start on e^x^3?
     
  2. jcsd
  3. May 24, 2014 #2

    micromass

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    You cannot find a solution to this integral in closed form using the usual elementary functions. You can prove this rigorously using the Liouville theorem.

    You can find an expression of the integral using the incomplete Gamma function or the exponential integral.
     
  4. May 24, 2014 #3
    But my textbook solutions has an answer for this...
    So originally, I was trying to solve the differential equation:
    y' + 3x^3y = x
    And then I found that the integrating factor was: e^∫3x^2 dx
    = e^x^3

    And my textbook said to use the integrating factor method.
     
  5. May 24, 2014 #4

    micromass

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  6. May 24, 2014 #5

    micromass

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  7. May 24, 2014 #6

    HallsofIvy

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    I presume you mean it was y'+ 3x^2y= x

    Okay, multiplying both sides of the equation by [itex]e^{x^3}[/itex] gives
    [itex](e^{x^3}y)'= xe^{x^3}[/itex] and, integrating,
    [itex]e^{x^3}y= \int_a^x te^{t^3}dt[/itex] where "a" gives the constant of integration.

    From that [itex]y(x)= e^{-x^3}\int_a^x te^{t^3}dt[/itex]

    That integral cannot be done in terms of elementary functions so either leave the solution as that or use the gamma function as wolfram alpha suggests.
     
  8. May 24, 2014 #7

    SammyS

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    Shouldn't that integrating factor be [itex]\displaystyle \ e^{\left(\Large \int 3x^3\,dx\right)}\ ?[/itex]
     
  9. May 24, 2014 #8

    HallsofIvy

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    Unless, as I suggested, the original differential equation was y'+ 3x^2y= x.
     
  10. May 24, 2014 #9

    Zondrina

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    You could also use the Taylor expansion, if the integral is indeed ##\int xe^{x^3} dx = \sum_{n=0}^{∞} \frac{1}{n!} \int x^{3n + 1} dx##.

    EDIT: You could also find values by using ##\int_{a}^{b}## instead of ##\int##.
     
    Last edited: May 24, 2014
  11. May 24, 2014 #10
    @Zondrina can you explain the taylor expansion method
     
  12. May 24, 2014 #11

    Zondrina

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    Using a Taylor expansion for ##e^x##, derive an expansion for ##e^{x^3}##. Now find one for ##xe^{x^3}## and then integrate it as desired.

    If you're interested in the reason, uniform continuity might be worth a read.
     
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