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Change of variable in indefinite integrals

  1. Oct 28, 2006 #1

    quasar987

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    I've had this very basic question on the back of my mind for 2 almost years now and I think I've found a satisfactory answer. The question is this most simple one: how do we justify a change of variables such as

    [tex]\int e^{ax}dx = \int \frac{1}{a}e^udu[/tex]

    in an indefinite integral? My "solution" is this... We are looking for a primitive [itex]F(\beta)[/itex] to [itex]f(\beta)[/itex]. By one version of the fondamental thm, as soon as f is continuous, such a primitive is given by

    [tex]F(\beta)=\int_{\phi(\alpha)}^{\phi(\beta)}f(x)dx = \int_{\alpha}^{\beta}(f\circ \phi)(s)\frac{d\phi}{ds}ds[/tex]

    where [itex]\phi:(\alpha,\beta)\rightarrow (\phi(\alpha), \phi(\beta))[/itex] is [itex]C^1[/itex]. (Ths is just the change of variable thm)

    This may be total nonsense, but my eyes are itching from tiredness so I wouldn't know. Any comments on the above or an opinion on how to justify change of variable in indefinite integrals will be warmly appreciate tomorrow morning.

    Good night!
     
  2. jcsd
  3. Oct 28, 2006 #2

    HallsofIvy

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    I'm not at all sure what you are asking. Certainly if
    [tex]F(x)= \int e^{ax}dx[/tex] and
    [tex]G(u)= \frac{1}{a}\int e^u du[/tex]
    then, by the chain rule, with u(x)= ax,
    [tex]\frac{dG}{dx}= \frac{dG}{du}\frac{du}{dx}[/tex]
    [tex]= \left(\frac{1}{a}e^u\right)(a)[/tex]
    [tex]= e^u= e^{ax}= \frac{dF}{dx}[/tex]
    Since F and G have the same derivatives they are anti-derivatives (indefinite integrals) of the same thing.

    "Substitution" in integrals is just the chain rule "in reverse".
     
  4. Oct 28, 2006 #3

    quasar987

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    ack, I had tried things with the chain rule, but never this. Yep, this is exactly what I was loking for, thanks Halls!
     
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