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Integral substitution fallacy

  1. Mar 22, 2015 #1
    Let ##x'=1/u' \Rightarrow dx' = \frac{-1}{u'^2} du'##. Then the integral ##\int_{x_0}^{x} x' dx'## turns into ##\int_{1/u_0}^{1/u} \frac{-1}{u'^3} du'##.

    Here comes the fallacy: ##\int_{1/u_0}^{1/u} \frac{-1}{u'^3} du' = [\frac{1}{2} \frac{1}{u'^2}]_{1/u_0}^{1/u} = \frac{1}{2} (u^2-u_0^2)##. Here I just substituted ##1/u## and ##1/u_0## into ##u'##, and I end up with getting something that is obviously wrong since ##\frac{1}{2} (u^2-u_0^2) = \frac{1}{2}(1/x^2 - 1/x_0^2) \neq \frac{1}{2}(x^2-x_0^2) ##.

    But why is that substitution incorrect? Or perhaps I set the limits wrong? Please help!
     
  2. jcsd
  3. Mar 22, 2015 #2

    Mark44

    Staff: Mentor

    Two things:
    1. It's very confusing to see the primes on your variables. I don't think x' denotes the derivative in your example. I could be mistaken, but if I'm not, using x' instead of x adds needless confusion.
    2. It's bad practice to use the same variable for a limit of integration and as the dummy variable for integration.
    If I simplify your example, I get no contradiction.
    Let x = 1/u, so dx = -du/u2

    Also, if x = a, then u = 1/a, and if x = b, then u = 1/b.
    Then ##\int_a^b x dx = -\int_{1/a}^{1/b} (1/u)(1/u^2)du##
    ##= -\int_{1/a}^{1/b} u^{-3}du = + (1/2) \left. u^{-2} \right|_{1/a}^{1/b} = (1/2) (b^2 - a^2)##

    If you do the integration with no substitution, you get
    ##\int_a^b x dx = \left. (1/2) x^2 \right|_a^b = (1/2) (b^2 - a^2)##
     
  4. Mar 22, 2015 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Using u'= 1/x' leads to a u' integral with limits 1/x and 1/x_0.
     
  5. Mar 22, 2015 #4
    aah, that's right. I screwed up while setting up the limits. Thank you :)
     
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