Definite vs. indefinite integrals for ei and erf

  • #1
CRGreathouse
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From a paper I've been reading:
This class includes, in addition to the elementary functions, a number of well-known special functions such as the exponential integral
[tex]\text{ei}(u)=\int\frac{u'}{u}e^u\,dx[/tex]
and the error function*
[tex]\text{erf}(u)=\int u'e^{u^2}\,dx[/tex]


* The usual error function, [itex]\text{Erf}(x)=\int_0^x\exp(-t^2)\,dt[/itex] [Bate53], differs from our definition, which is denoted as Erfi in [Bate53], as follows: [itex]\text{Erf}(x)=1/i\text{Erfi}(ix).[/itex] Also see the Appendix.
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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  • #2
tiny-tim
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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?)
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
 
  • #3
CRGreathouse
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Isn't u' just du/dx, where erf is a function of u, and u is a function of x?
But u isn't a function! It's just the argument of the function.
 
  • #4
D H
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But u isn't a function! It's just the argument of the function.
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!
 
  • #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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