$f$ is integrable but has no indefinite integral Let $$f(x)=\cases{0,& $x\ne0$\cr 1, &$x=0.$}$$
Then $f$ is clearly integrable, yet has no antiderivative, on any interval containing $0,$ since any such antiderivative would have a constant value on each side of $0$ and have slope $1$ at $0$—an impossibility.
So does this mean that $f$ has no indefinite integral?
EDIT
My understanding is that the indefinite integral of $f$ is the family of all the antiderivatives of $f,$ and conceptually requires some antiderivative to be defined on the entire domain. Is this correct?

Just to supplement Emanuele’s answer. If $ I $ is an open subset of $ \mathbb{R} $, then some mathematicians define the indefinite integral of a function $ f: I \to \mathbb{R} $ as follows:
$$
\int f ~ d{x} \stackrel{\text{def}}{=} \{ g \in {D^{1}}(I) ~|~ f = g' \}.
$$
Hence, taking the indefinite integral of $ f $ yields a family of antiderivatives of $ f $. If $ f $ has no antiderivative, then according to the definition above, we have
$$
\int f ~ d{x} = \varnothing.
$$
