$f$ 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?
 A: This is a matter of definitions. Usually an antiderivative of a function $f$ is any function whose derivative is $f$. The indefinite integral usually denotes the set of all antiderivatives.
Since every derivative satisfies the intermediate values property, your function $f$ cannot be a derivative and hence has no indefinite integral.
A different thing is the integral function which might be defined as a function $F$ such that 
$$
  \int_a^b f(x) dx = F(b)-F(a).
$$
This exists for all integrable functions.
The fundamental theorem of Calculus states that the two concepts agree for continuous functions.
A: 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
$$
\int f ~ d{x} = \varnothing.
$$
A: Attempting to express $f$'s indefinite integral on its entire domain $\mathbb R$ runs into a technical problem at $x=0:$ $$\int f(x)\,\mathrm dx =\cases{C_1,& $x<0$\cr \mathbf{???}, &$x=0$\cr C_2,& $x>0$}.$$
No indefinite integral $\iff$ no antiderivative.
On the other hand, $f$ has an indefinite integral (and antiderivative) on every interval not containing $0.$
