Darboux' theorem says that, if $f\colon \mathbb{R} \rightarrow \mathbb{R}$ has an antiderivative, than $f$ has to satisfy the intermediate value property, namely, for any $a<b$ , for any number $C$ with $f(a)<C<f(b)$ or $f(b)<C<f(a)$ , there exists a $c \in (a,b)$ such that $f(c) = C$ . With this theorem, we understand that if $f$ does not satisfy the intermediate value property, then no function $F$ satisfies $F' = f$ on $\mathbb{R}$ .

Now, we will give an example to show that the converse is not true, i.e., a function that satisfies the intermediate value property might still have no antiderivative.

Let

First let us see that $f$ satisfies the intermediate value property. Let $a<b$ . If $0<a$ or $b\leq 0$ , the property is satisfied, since $f$ is continuous on $(-\infty,0]$ and $(0,\infty)$ . If $a\leq 0<b$ , we have $f(a) = 0$ and $f(b) = (1/b)\cos(\ln b)$ . Let $C$ be between $f(a)$ and $(b)$ . Let $a_0 = \exp(-2\pi k_0 +\pi)$ for some $k_0$ large enough such that $a_0 < b$ . Then $f(a_0)=0 = f(a)$ , and since $f$ is continuous on $(a_0,b)$ , we must have a $c \in (a_0,b)$ with $f(c) = C$ .

Assume, for a contradiction that there exists a differentiable function $F$ such that $F'(x) = f(x)$ on $\mathbb{R}$ . Then consider the function $G(x) = \sin(\ln x)$ which is defined on $(0,\infty)$ . We have $G'(x) = f(x)$ on $(0,\infty)$ , and since it is a an open connected set, we must have $F(x) = G(x) + c$ on $(0,\infty)$ for some $c\in\mathbb R$ . But then, we have