Function that has no antiderivative

Question

Can someone give an example of a function $f:$ $]-1,1[ \rightarrow \mathbb{R}$ that has no antiderivative? Is there a certain "class" of functions which share this characteristic? Thank you in advance.

what does that notation mean? is (-1,1) the domain or range – Saketh Malyala Jun 18 '17 at 21:11 — Saketh Malyala, Jun 18 '17 at 21:11
Related: https://math.stackexchange.com/q/239324/120540 – pjs36 Jun 18 '17 at 21:15 — pjs36, Jun 18 '17 at 21:15

score 3 · Accepted Answer · answered Jun 18 '17 at 21:41

$\DeclareMathOperator{sgn}{sgn}$As the previous answer suggests, the good example is the signum function. If $F'(x)=\sgn(x)$ for any $x$, then $F'(x)=1$ if $x>0$, so then $F(x)=x+C$. Similarly for $x<0$ we have $F(x)=-x+D$. Of course, $F$ should be continuous at $0$. Because $F(0+)=C$ and $F(0-)=D$, we get $F(0)=C=D$. Hence $F(x)=|x|+C$ for any $x$. But $F$ is not differentiable and it could not be an antiderivative.

score 1 · Answer 2 · answered Jun 18 '17 at 21:20

1

Any function that does not have the intermediate value property does not have an antiderivative.

This can be seen as an immediate consequence of Darboux theorem, but I am sure there are more "fundamental" ways to prove it

answered Jun 18 '17 at 21:20

Ant

21,098

Isn't this claim just a paraphasing of Darboux's theotrem itself? If it had a more elementary proof, then that proof would also prove Darboux itself. – hmakholm left over Monica Jun 18 '17 at 21:57
@HenningMakholm Good point. – Ant Jun 19 '17 at 06:08

Function that has no antiderivative

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