Can a function $f$ have an antiderivative even though its indefinite integral $F(x) = \int_{a}^{x} f(t)\ dt$ is not one? The fundamental theorem of calculus states that if $f:[a,b] \to \mathbb{R}$ is integrable and $F(x) = \int_{a}^{x} f(t)\ dt$, then $F'(x) = f(x)$ at every point $x$ at which $f$ is continuous. This means that if $f$ is integrable, $F'(x) = f(x)$ almost everywhere. If $f$ is continuous on $[a,b]$, then $F$ is an antiderivative of $f$, since $F'(x) = f(x)$ holds for all $x \in [a,b]$. 
But what if $f$ has a discontinuity at some $x \in [a,b]$? In this case, it is not necessarily true that $F'(x) = f(x)$, and so we cannot necessarily conclude that $F$ is an antiderivative of $f$. Does this mean that there is no antiderivative of $f$? Is it possible for $f$ to have an antiderivative but the indefinite integral $F$ is not an antiderivative of $f$?
I know that if $f$ has a jump discontinuity, then $f$ can have no antiderivative (since the derivative of a function must satisfy the intermediate value property), but what if we have some other type of discontinuity?

Consider the map$$\begin{array}{rccc}F\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $F$ is differentiable and$$(\forall x\in\Bbb R):F'(x)=\begin{cases}-\cos\left(\frac1x\right)+2x\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}$$So, $F'$ is discontinuous at $0$. But $F$ is an antiderivative of $F'$.
