# Definite vs. indefinite integrals for ei and erf

Homework Helper
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
$$\text{ei}(u)=\int\frac{u'}{u}e^u\,dx$$
and the error function*
$$\text{erf}(u)=\int u'e^{u^2}\,dx$$

* The usual error function, $\text{Erf}(x)=\int_0^x\exp(-t^2)\,dt$ [Bate53], differs from our definition, which is denoted as Erfi in [Bate53], as follows: $\text{Erf}(x)=1/i\text{Erfi}(ix).$ 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.

tiny-tim
Homework Helper
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! 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

Homework Helper
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.

D H
Staff Emeritus
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, $u'dx = du$ 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!