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A very Weird Integral

  1. Jan 27, 2012 #1
    1. The problem statement, all variables and given/known data

    evaluate 5}\right)dx%20$.gif

    2. Relevant equations

    3. The attempt at a solution
    i dont think there is an elementary function as anti derivative for this integral

    i tried taylor expansion, doesn't seem to work.

    Can anyone give me a hint ?
  2. jcsd
  3. Jan 27, 2012 #2


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    WolframAlpha shows the indefinite integral to include the Hypergeometric function 2F1. http://www.wolframalpha.com/input/?i=Integrate+%281-x^5%29^%281%2F7%29-%281-x^7%29^%281%2F5%29dx

    It shows the definite integral to be zero.
  4. Jan 27, 2012 #3
    Although I really don't think there's an easy way to solve this one, it's neat to notice the weird symmetry of the powers on x and of the roots (1/5 and 1/7 compared with 7 and 5 for powers). Notice that you can factor both (1-x^5) and (1-x^7). It's kind of taunting when you do factor them and realize they're very similar. Try playing around with this.
  5. Jan 27, 2012 #4


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    There's a neat way to do this. Start by observing that:

    [itex]f(x) = (1-x^5)^{\frac{1}{7}}[/itex] and [itex]g(x) = (1-x^7)^{\frac{1}{5}}[/itex]

    are in fact inverse functions. This means that they are symmetrical when reflected in the line y = x.

    Over the domain [0,1], they both have the range [0,1].

    Can you now draw a conclusion about the areas under each curve from 0 to 1? :wink:

    This general method applies to a lot of functions of that form. Nothing special about the 5 or 7 here. :biggrin:
  6. Jan 27, 2012 #5
    Bah, I was getting there! Nice find Curious.
  7. Jan 27, 2012 #6
    Wow thank you and everyone.
    mathematics is awesome
  8. Jan 27, 2012 #7


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    Yes, this is very interesting!

    So, under what conditions is [itex]\displaystyle \int_0^1 \left(f(x)-f^{-1}(x)\right)\,dx=0\,?[/itex]

    Of course the domains & ranges have to all be [0,1].

    As it so happens, map 1 → 0 , and 0 → 1 . Is that important ? Yes it is.

    In the case where f(0) = 0 = f -1(0) and f(1) = 1 = f -1(1), we find that [itex]\displaystyle \int_0^1 \left(f(x)+f^{-1}(x)\right)\,dx=1[/itex] instead. Notice that the functions are added here.
  9. Jan 27, 2012 #8


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    Yes, it depends on whether the functions are increasing or decreasing over the interval. :smile:
  10. Jun 27, 2012 #9
    You can also follow the hints that you were given: first substitute, say in the second integral, [itex] t=(1-x^7)^{1/5} [/itex]. You'll get something like [itex] \int dt t f(t) [/itex]. Then do partial integration with this. Since the integration limits are 0 and 1 in both cases, the remaining integrals should cancel and you're left with a boundary term which is easy to evaluate.
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