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Homework Help: Conditions for the applicability of u-substitution

  1. Oct 13, 2018 #1
    What are the conditions for applicability of u-substitution, i.e. when does it not work? Note that I'm not asking when is it a bad idea (that won't get you any closer to evaluating the integral), but are there any conditions that cause u-sub to yield wrong answers?

    I started running into what I think is a case of u-sub not working when I was thinking about integrals of odd functions, ex:

    $$\int_{-\infty}^{\infty}x e^{-a x^2}dx$$
    $$u=x^2$$
    $$du=2xdx$$
    $$\int_{-b}^{b}x e^{-a x^2}dx=\frac{1}{2}\int_{b^2}^{b^2} e^{-a u}dx=0$$
    Where the last part equals zero because now the bounds are equal...
    But this begged the question... why can't I just u-sub in such a way that the bounds on the integral are always equal, and all integrals go to 0? - Obviously there has to be some constraint on the applicability of u-sub.

    Here is an example where it seems u-sub just leads to the wrong answer...

    $$\int_{-2}^{1}x^4dx=\frac{33}{5}=6.6$$
    With $$u=x^2$$ we get:

    $$\frac{1}{2}\int_{4}^{1}u^{3/2}du=-\frac{31}{5}=-6.2$$

    So what gives? What basic mathematical principle is being violated here? I'm sure I learned this at some point... waaay back when.
     
  2. jcsd
  3. Oct 14, 2018 #2

    Charles Link

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    It's interesting. I haven't studied it in sufficient detail to give a complete answer, but one thing that stands out is that you don't have a one-to-one continuous mapping from ## x## to ## u ##. I think that is likely to be the source of the difficulty.
     
  4. Oct 14, 2018 #3

    PeroK

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    If ##u = x^2##, then ##x = \pm \sqrt{u} = \pm u^{1/2}##. In particular, the equation ##x = u^{1/2}## is not correct for all ##x## in this case.

    You have, therefore, a different relationship between ##x## and ##u## depending on whether ##x## is positive or negative. So, you need to split the integral to do the substitution in this case.
     
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