Definite vs. indefinite integrals for ei and erf

[CRGreathouse]

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]

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

 

[CRGreathouse]

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]

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!

 

[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!