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Integral / Complex Nos.

  1. Feb 17, 2005 #1
    [tex]\int _1 ^9 \frac{1}{\sqrt[3]{x-9}}\: dx = -6\cos \left( \frac{2k\pi}{3} \right) - i \mbox{ } 6\sin \left( \frac{2k\pi}{3} \right) \quad k \in \mathbb{N}[/tex]

    How do I interpret this integral? The only real number that I can get out of it is -6.


    PS: I mean, maybe the integral can't be interpreted as the area under the graph anymore. What happens in this case? I'm a bit confused. :smile:
    Last edited: Feb 17, 2005
  2. jcsd
  3. Feb 17, 2005 #2
    [tex]\int _1 ^9 \frac{1}{\sqrt[3]{x-9}} dx = \int _1 ^9 \frac{1}{\sqrt[3]{(-1)(9-x)}} dx = \frac{1}{\sqrt[3]{-1}} \int _1 ^9 \frac{1}{\sqrt[3]{9-x}} dx [/tex]

    now, it become a real integral multiply by a complex number..

    since [itex] \sqrt[3]{-1} = e^{i(2\pi k+\pi)/3} [/tex], k=0,1,2

    [tex] \frac{1}{\sqrt[3]{-1}} = e^{-i(2\pi k+\pi)/3} [/tex]

    [tex] \int _1 ^9 \frac{1}{\sqrt[3]{9-x}} dx = 6 [/tex] as you suggest

    therefore, the answer is :
    [tex] 6e^{-i(2\pi k+\pi)/3} [/tex], k=0,1,2
    [tex] 6 (cos(\pi/3)-isin(\pi/3))= 3 ( 1-i\sqrt{3} )[/tex]
    [tex] 6 (cos(3\pi/3)-isin(3\pi/3))=-6[/tex]
    [tex] 6 (cos(5\pi/3)-isin(5\pi/3))=3 ( 1+i\sqrt{3} )[/tex]
  4. Feb 17, 2005 #3
    Oh, I see what you mean. The given integral can still be interpreted as an area (after some rearrangements) that is multiplied by a complex number, making the result complex.

  5. Feb 17, 2005 #4


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    Make the substitution [itex] 9-x=u [/itex] and solve the integral.

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