$f$ 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?

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 $0.$
