A integral similar to Cauchy's integral formula Let $f$ be holomorphic on an open set containing the closed unit disc, and $\gamma$ be the unit circle, parametrized counterclockwise. Prove that for any $a \in \mathbb{C},$ $|a| \neq 1$,
$$
\frac{1}{2 \pi i} \int_{\gamma} \overline{f(z)}/(z-a) dz= 
\begin{cases} \overline{f(0)} &\mbox{if} \,\, |a|<1, \\
              \overline{f(0)}-\overline{f(1/ \bar{a})} &\mbox{if}\,\, |a|>1.
\end{cases}
$$
I don't know how to deal with the integral since $\overline{f}$ is not holomorphic. I was trying to use Mean Value Theorem of harmonic function but still not working. Any thoughts are appreciated.

I suggest you think about the following approach. The problem with conjugation, as you have observed already, is that it is not holomorphic, but even though $\overline{f(z)}$ is not a holomorphic function the "double" conjugate $f^*(z) := \overline{f(\overline{z})}$ is holomorphic.  (Indeed, if the power series of $f(z)$ around a point has coefficients $c_n$, the function $f^*(z)$ has a power series around the same point with coefficients $\overline{c_n}$; this was an exercise when I took my first complex analysis course.) So you'd be better off if somehow you could work with $\overline{f(\overline{z})}$ instead of $\overline{f(z)}$. But you don't want to make the change of variables $w = \overline{z}$ because that is not holomorphic. 
Key idea: on the unit circle, which is the path of integration, we have $\overline{z} = 1/z$ and the function $1/z$ is holomorphic.  So make the change of variables $$w = 1/z$$ in the integral. Be careful about the path of integration, which remains the unit circle but gets reversed orientation, so you'll have some sign issues to deal with. In the integrand the numerator will be $\overline{f(1/w)} = \overline{f(\overline{w})} = f^*(w)$ (we have $1/w = \overline{w}$ because the path of integration is the unit circle) and the denominator will be a quadratic polynomial in $w$ that you can factor and split up into a sum of two terms using partial fractions. You can  apply the usual Cauchy integral formula to the two integrals, taking cases on one of them if $|a| < 1$ or $|a| > 1$. I tried this myself and everything worked, but I will leave the detailed calculations to you because it is a homework problem and I think it will be instructive to do it on your own.
