# Definite vs. indefinite integrals for ei and erf

• CRGreathouse
In summary, the given conversation discusses the notation and interpretation of special functions, such as the exponential integral and error function, which are defined with definite integrals but can also be expressed with indefinite integrals using substitution. The conversation also mentions some confusion regarding the use of u as both the argument and variable in these functions.

#### 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
$$\text{ei}(u)=\int\frac{u'}{u}e^u\,dx$$
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.

CRGreathouse said:
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

tiny-tim said:
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.

CRGreathouse said:
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!

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!