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Topic:
why do not CAS see that int( f'(x) ,x) = f(x) ?
Replies:
19
Last Post:
Nov 18, 2017 12:46 AM




why do not CAS see that int( f'(x) ,x) = f(x) ?
Posted:
Nov 10, 2017 10:24 PM


This question came up in Mathematica forum.
https://mathematica.stackexchange.com/questions/159688/differentiatingandintegratingsimplefunctionslikefxlogsinxlogc
I tried on Maple also. The question is basically saying, when don't CAS (Mathematica in the question) "see" right away that
int( f'(x),x ) = f(x)
It seems to me what happens is that Maple and Mathematica first evaluate f'(x), and then spend long time trying to integrate the result, coming up with very complicated antiderivative.
I run FullSimplify[] on the antiderivative trying to see if will give back the original function f(x) and its been running for hours.
Only Fricas is smart enough to do this. (unless I made mistake in the input)Below is what I tried
Mathematica: ============= ClearAll[f,x]; f[x_]:=Log[Sin[x]]Log[Cos[x]]; sol=Integrate[f'[x],x] ... VERY complicated result ....
Maple ======= restart; f:=x>log(sin(x))*log(cos(x)); int(diff(f(x),x),x); ... VERY complicated result, like Mathematica...
FriCAS ======= http://axiomwiki.newsynthesis.org/FriCASIntegration#bottom
\begin{axiom} setSimplifyDenomsFlag(true) f:=log(sin(x))*log(cos(x)) integrate(D(f,x),x) \end{axiom} log(sin(x))*log(cos(x))
So my question is: Why M and Maple do not see right away that int(f'(x),x) = f(x) ? isn't this what FTOC says?
Is it just an issue of parsing? I.e. they will first evaluate f'(x) and then try to integrate the result, and they just need to add extra code to detect that the integrand is a derivative of some function and do the shortcircuit FTOC to return that function asthe answer right away?
I assume this is what FriCAS did.
Or is there more to it?
Nasser



