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Definite vs. indefinite integrals for ei and erf

  1. May 13, 2009 #1

    CRGreathouse

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    From a paper I've been reading:
    I wanted to know how to interpret the notation. What does it mean to take the derivative of u when it is a function argument? How do I take the integral wrt x of a seemingly-constant expression and not end up with x in the answer? Generally, how are these functions (typically defined with definite integrals) defined with indefinite integrals? (Or is this just a trick of notation?)

    I'd be happy to post more of the paper if it would help. I'm actually recopying it at the moment, so it wouldn't be too hard.
     
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  3. May 14, 2009 #2

    tiny-tim

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    Hi CRGreathouse! :smile:

    Isn't u' just du/dx, where erf is a function of u, and u is a function of x?

    So eg if u(x) = ix, then u' = i and erf(u) = ∫ie-x2dx
     
  4. May 14, 2009 #3

    CRGreathouse

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    But u isn't a function! It's just the argument of the function.
     
  5. May 14, 2009 #4

    D H

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    Who says it's not a function? Assume it is, and apply the freshman calculus substitution rule. Using informal notation, [itex]u'dx = du[/itex] and thus

    [tex]\int \frac{u(x)'}{u(x)}e^{u(x)}\,dx = \int \frac{e^{u(x)}}{u(x)} u'(x)dx = \int \frac{e^u}{u} du = Ei(u)[/tex]

    The last is a bit of abuse of notation as the exponential integral is a definite integral, not an indefinite integral.

    You've been doing math so long you forgot the basics!
     
  6. May 14, 2009 #5

    CRGreathouse

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    Thanks, DH. Between the bound-unbound abuse of notation (u as argument and running variable) and the indef/def abuse of notation I wasn't sure how to interpret it!
     
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