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Integration of (x^2-x^3)^-(1/3) from 0 to 1

  1. Apr 14, 2005 #1


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    Hey all,

    I need some help with a multivalued complex integration problem.

    Evaluate the integral of (x^2-x^3)^-(1/3) from 0 to 1.

    I know you need to pick branches of the cube root function, and this will give you multiples of the integral which you can then equate to integrals along contours around the singularities 0 and 1. However, I'm having trouble finding those integrals around the contours. How can I use residues when fractional powers are involved? Any help is appreciated.
  2. jcsd
  3. Apr 14, 2005 #2


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    Here's the indefinite integral

    [tex] \int \frac{dx}{\sqrt[3]{x^{2}-x^{3}}} =\frac{1}{\sqrt[3]{x^{2}-x^{3}}}\left[\left(3x\sqrt[3]{1-x}\right) \ _{2}F_{1}\left(\frac{1}{3},\frac{1}{3},\frac{4}{3},x\right)\right] +\mathcal{C} [/tex]

    Good luck with those limits (in case u use the FTC).

    Last edited: Apr 14, 2005
  4. Apr 14, 2005 #3
    Do you actually have a methodology to get solutions in the form of hypergeometric polynomials or did you just plug that into Mathematica or Maple?

    I mean either of those programs will tell you what the limit is too (at least maple does). But if there's a method to get that answer I'd be curious how you go about it. Always nice to have a new method of integration in your arsenal.

  5. Apr 15, 2005 #4
    Well, I solved this in time to turn it in for class. If anyone's interested, the method is to integrate the function on a barbell shaped contour around the singularities 0 and 1. Then the curved paths around the singularities go to 0 by limiting arguments, and the straight paths give (1-e^(2 pi i / 3)) I, where I is the desired integral, by branch arguments. Deforming the contour to infinity and using limiting arguments gives the integral around the contour also equals -2 pi i e^(pi i / 3). Equating the two and solving for I gives I = pi / sin (pi / 3), or 2 pi / sqrt(3).

    By the way, this is the screenname i usually use...
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