0
$\begingroup$

My math books, in the introductory chapter of indefinite integrals (they call them primitives, and a primitive of a function is any function who's derivative is the original function) concludes the following thing: Any continuous function on the reals admits primitives on it's domain (that means that there exist indefinite integrals).

However I can easily come up with a counter-example. The function $f(x) = x^x$ is continuous on $[0,1]$ however there exists no such function $F(x)$ such that $F'(x) = x^x$.

Did I misinterpret the conclusion?

  • 3
    There exists such an $F$! You can take $F(x) = \int_0^x t^t\, dt$, for example.2012-08-13
  • 1
    What about $F(x) = \int_0^x t^t dt$?2012-08-13
  • 3
    (Blargh, too slow; too few martinis today.)2012-08-13
  • 0
    Probably the counter-example do not want to say "there exists no such function", but "cannot be expressed in terms of elementary functions".2012-08-13
  • 0
    Besides the two comments above, one question: are you pitch sure your function's continuous at $\,x=0\,$? If so, what is $\,0^0\,$?2012-08-13
  • 0
    Ah, it seems that the book wasn't clear enoug. Yes there are functions that satisfy those conditions, but what I was reffering to was that there is no way of expressing it in elementary terms. And isn't $0^0 = 0$ and if not, then I can ask about $[0.5, 1]$ there still isn't an elementary way to express it.2012-08-13
  • 0
    You can extend $x^x$ by continuity in $0$, by setting $f(0)=1$.2012-08-13
  • 0
    Indeed @enzotib...as long as $\,x>0\,$ ,which is what the OP needs, I believe.2012-08-13
  • 0
    As already shown, $x^x$ can be integrated on a suitable interval; we just don't know how to express it in terms of more familiar functions.2012-08-13

1 Answers 1

0

Your book deserves a bit of criticism, in my humble opinion. The best way to go in a safe way is to assume that you are always working with functions defined on an interval. Indeed, it may be troublesome to find a primitive of the continuous function $f \colon \mathbb{Z} \to \mathbb{R}$ defined by $f(n)=n$. You would need to define the primitive on an open set that contains $\mathbb{Z}$, but then you'd lose uniqueness up to a constant.

However, if $f$ is defined on some interval $I$ and you look for a function $F \colon I \to \mathbb{R}$ such that $F'=f$ everywhere in $I$, then the continuity of $F$ in $I$ is a sufficient condition for $F$ to exist.