How much can the hypotheses of the Schwarz integral formula be weakened? According to https://en.wikipedia.org/wiki/Schwarz_integral_formula, which I will paraphrase here, if $f$ is a holomorphic on the closed unit disk $\{ \lvert z \rvert \leq 1 \}$, then
$f(z) = \frac{1}{2\pi i} \oint_{\lvert \zeta \rvert = 1} \frac{\zeta+z}{\zeta-z}\operatorname{Re}(f(\zeta)) \frac{d\zeta}{\zeta} + i \operatorname{Im}(f(0))$,
for all $\lvert z \rvert < 1$.
My question is this, how much can one weaken the hypothesis that $f$ is holomorphic on $\overline{D}$, while retaining the conclusion?
For instance, if $f$ is only assumed to be holomorphic on the open disk $D$, and continuous on $\overline{D}$, will the conclusion hold? If not, can one add an extra hypothesis, so that the conclusion will hold, and will allow the function to have some kind of singularity at the boundary. The type of singularities I am interested in are branch type singularities, such as fractional powers $(z-1)^a$, say, for some $a>0$.

If $f$ is holomorphic in the open disk and continuous on the closed disk the formula is valid. All you have to do is to apply the formula for the disk $|z|=1-\epsilon$ and then  let $\epsilon \to 0$. 
