# Definite vs. indefinite integrals for ei and erf

1. May 13, 2009

### CRGreathouse

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.

2. May 14, 2009

### tiny-tim

Hi CRGreathouse!

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. May 14, 2009

### CRGreathouse

But u isn't a function! It's just the argument of the function.

4. May 14, 2009

### D H

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

$$\int \frac{u(x)'}{u(x)}e^{u(x)}\,dx = \int \frac{e^{u(x)}}{u(x)} u'(x)dx = \int \frac{e^u}{u} du = Ei(u)$$

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. May 14, 2009

### CRGreathouse

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!