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comp.soft-sys.math.mathematica |
(*snip*) > q=Exp[-(x1-t)^2/2*sigma^2*t] > Integrate[q, {t, .5,1}] > Now Mathematica does not solve this integral, it just repeats the > I am trying to get an expression in terms of x1. Why do I get a It is conventional (by design) that Mathematica returns an expression For instance, having started a new Mathematica session, if we try to In[1]:= f[2] Out[1]= f[2] Now, we give a definition (a meaning) to the symbol f. In[2]:= f[x_] = 2 x Out[2]= 2 x From now on, evaluating f will return a value. In[3]:= f[2] Out[3]= 4 Of course, *Integrate* is a built-in function that has already a For instance, Mathematica knows how to integrate E^(-x^2) (in terms of In[1]:= Integrate[Exp[-x^2], x] Out[1]= 1/2 Sqrt[\[Pi]] Erf[x] In[2]:= Integrate[Exp[-x^3], x] Out[2]= -((x Gamma[1/3, x^3])/(3 (x^3)^(1/3))) In[3]:= Integrate[Exp[-x^3 - x^2], x] Out[3]= Integrate[E^(-x^2 - x^3), x] Regards,
> command
> statement like this instead of an answer?
unevaluated when Mathematica does not know how to evaluate this
expression. This can happen for user-defined functions as well as
built-in functions (though in special circumstances).
evaluate f[2], Mathematica just returns f[2] since it has not the
slightest idea of what the function f can possibly do.
meaning. Still, if Mathematica does not know how to find a definite or
indefinite integral, it returns the original expression as answer.
error function) and E^(-x^3) (in terms of gamma function) but not
E^(-x^3 - x^2) (the expression is returned unevaluated).
Jean-Marc