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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?

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There exists such an $F$! You can take $F(x) = \int_0^x t^t\, dt$, for example. –  martini Aug 13 '12 at 15:02
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What about $F(x) = \int_0^x t^t dt$? –  M.B. Aug 13 '12 at 15:02
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(Blargh, too slow; too few martinis today.) –  M.B. Aug 13 '12 at 15:04
    
Probably the counter-example do not want to say "there exists no such function", but "cannot be expressed in terms of elementary functions". –  enzotib Aug 13 '12 at 15:04
    
Besides the two comments above, one question: are you pitch sure your function's continuous at $\,x=0\,$? If so, what is $\,0^0\,$? –  DonAntonio Aug 13 '12 at 15:05

1 Answer 1

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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.

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