I'm guessing you want something like:
(%i1) erf_integral_rep(e) := subst('erf = lambda([s], block([z : ?gensym()], (2/sqrt(%pi)) * 'integrate(exp(-z^2),z,0,s))),e)$
(%i2) erf_integral_rep(erf(z) = 42);
(%o2) (2*integrate(%e^(-g34177^2),g34177,0,z))/sqrt(%pi)=42
I doubt that this will work all that well for you--Maxima evaluation reverts the work of erf_integral_rep:
(%i3) ev(%,nouns);
"Is "z" positive, negative, or zero?"pos;
(%o3) erf(z)=42
The CL gensym is obnoxious---the next Maxima will have a better way, I think.
--Barton
-----maxima-bounces at math.utexas.edu wrote: -----
>To:?maxima at math.utexas.edu
>From:?Julien?Martin?<balteo at gmail.com>
>Sent?by:?maxima-bounces at math.utexas.edu
>Date:?07/27/2010?04:48AM
>Subject:?[Maxima]?evaluation?of?the?erf?function
>
>Hello,
>
>I?get?the?following?result?with?Maxima:
>
>f(x)=(%e^(-(%theta*x^2)/%sigma^2)*(sqrt(%pi)*%k1*erf((sqrt(-%theta)*x)/%si
>gma)+2*%k2*%sigma*sqrt(-%theta)))/(2*%sigma*sqrt(-%theta))
>
>Is?there?no?way?for?Maxima?to?evaluate?the?value?of?the?erf?function
>directly?instead?
>
>
>It?would?then?replace?erf(x)?with?the?evaluated?version?of?2/sqrt(%pi)
>integral?from?0?to?x?of?exp(-t^2)?dt...
>
>How?do?I?enable?this?by?default?for?all?my?results?
>
>Thanks,
>
>Julien.
>
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