Let $\mu$ be a complex regular Borel measure on $\mathbb T.$ Define a function $u : \mathbb D \longrightarrow \mathbb C$ by $$u \left (re^{i \theta} \right ) = \int_{\mathbb T} \frac {1 - r^2} {1 + r^2 - 2r \cos (\theta - t)}\ d\mu \left (e^{it} \right ),\ 0 \leq r \lt 1,\ 0 \leq \theta \lt 2 \pi.$$
Show that $u$ is a harmonic function.
One way to do that is to interchange the Laplacian and the integral operator and then figure out whether the integrand is harmonic or not. Other way is to use the tricks in complex analysis to see whether $u$ is the real or imaginary part of an analytic function. So in order to apply the first trick in some sense we would like to have a sufficient condition for which the Laplacian operator can be brought inside the integral sign. For the second trick we need to find a suitable analytic function such that one of whose components is the function $u$ we have started with. Unfortunately none of the approach seems easy to me. Could anyone give me some suggestion in this regard?
Thanks in advance.
For each fixed $z \in \mathbb D$ we can define a new complex measure $\nu_z$ as $d\nu_z = (e^{it}+z)\ d\mu.$ Then the said integral is the real part of the following integral $$u(z) = \int_{\mathbb T} \frac {d\nu_z(e^{it})}{e^{it}-z}$$ and it is easy to show that given $a\in \mathbb D$ and $r\gt 0$ with $B(a,r) \subseteq \mathbb D$ the infinite series $\sum\limits_{n= 0}^{\infty} \frac {(z-a)^n}{(e^{it}-a)^{n+1}}$ converges uniformly to the integrand $\frac {1} {e^{it} - z}$ on $\mathbb T$ for each fixed $z \in B(a,r)$ and hence $u(z)$ admits a power series expansion about $a$ on $B(a,r)$ and therefore $u$ is holomorphic on $\mathbb D$ (since $a \in \mathbb D$ was arbitrary). Since real part of a holomorphic function is harmonic the result follows.