Definite integration yields a strange answer. (Solved)
11-14-2015, 05:17 PM
Post: #12
 parisse Senior Member Posts: 1,051 Joined: Dec 2013
RE: Definite integration yields a strange answer. (Sufficiently Explained - Updated)
The rule you show probably assumes integer exponents (because of the choice of the letter n) and it is perfectly adapted to highschool math. The rule is also valid for real exponents, provided that x is >0, because otherwise x^alpha is not defined since x^alpha:=exp(alpha*ln(x)).
Defining fractional powers x^(p/q) for x>0 can be done without introducing logarithms by solving a^q=x^p, and this can be extended to x<0 if q is odd, but the general convention used by CAS is not that one (probably because it is not defined for q even), it is take the principal determination of the complex logarithm ln(-x)+i*pi and apply exp(p/q*(ln(-x)+i*pi)). Numerically speaking x^(p/q) is also computed by applying the rule exp(p/q*ln(x)).
The Prime provides two functions: x^(p/q) is the normal complex definition and NTHROOT (also named surd in other CAS like giac or maple) remains in the real domain (if q is odd) and I believe it's the best choice one can do.
What I do not understand is that it seems (from what I read in the forums) that there are exercices in the US textbooks with definite integrations with fractional powers of x with x negative, using the x^(p/q) notation instead of the n-throot notation. I don't understand that because it does not make you understand fractional powers or integration better than you would if x was positive. To the contrary, it will confuse people.
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