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Definite integral: why is this wrong?

  1. Feb 1, 2013 #1
    According to wolfram alpha, the correct answer is -3/4.
     

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  3. Feb 1, 2013 #2
    Never mind. It is right.
     
  4. Feb 1, 2013 #3
    Wolfram isn't getting 3/4, it's getting (-3/2)(-1)^(2/3) which it is computing to be approximately equal to 0.75-1.29904i

    The real question of the day is why (-1)^(2/3) doesn't evaluate to 1. :uhh:
     
  5. Feb 1, 2013 #4

    jbunniii

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    There are three valid answers. Wolfram is calculating this one:
    $$(-1)^{2/3} = (e^{i\pi})^{2/3} = e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -1/2 + i\sqrt{3}/2$$
    These two also work:
    $$(-1)^{2/3} = (e^{-i\pi})^{2/3} = e^{-i2\pi/3} = \cos(-2\pi/3) + i\sin(-2\pi/3) = -1/2 - i\sqrt{3}/2$$
    $$(-1)^{2/3} = (e^{i3\pi})^{2/3} = e^{i2\pi} = 1$$
     
  6. Feb 1, 2013 #5

    Ray Vickson

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    There are other valid answers as well. If we put ## \sqrt[3]{x-1} = -\sqrt[3]{1-x} \; (0 \leq x \leq 1)## we have
    [tex] \text{answer } = -\int_0^1 \frac{1}{\sqrt[3]{1-x}}\, dx = -3/2.[/tex]
     
  7. Feb 1, 2013 #6
    Mind has now been totally blown.
     
  8. Feb 2, 2013 #7

    Dick

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    Oh, I wouldn't blow that hard. If you are doing this as a real integral then you want to pick the branch of cube root which is always real. Your original answer was fine. Wolfram just picked the wrong branch. You could also with equal justification claim that the integral from 0 to 1 of x^(1/3) has three possible answers. Or even more if you get more creative. But that's certainly not the usual answer. Wolfram just misinterpreted the question.
     
    Last edited: Feb 2, 2013
  9. Feb 2, 2013 #8

    SammyS

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