This is an exercise from Ahlfors' Complex Analysis text.
I need to show that the mean value property holds for the function $u=\log|1+z|$ in the circle with center $z_0=0$ and radius $r=1$.
The function is harmonic in the open disk $B(0;1)$ so clearly the integrals $\int_0^{2\pi}\log|1+re^{i \theta}| \mathrm{d} \theta$ are all zero (the value of $u$ at the origion) for $0\leq r <1$.
I tried taking the limit $r \to 1^-$ using some convergence theorem, but I failed.
How can you show that the (improper) integral $\int_0^{2\pi}\log|1+e^{i \theta}| \mathrm{d} \theta$ is zero as well?
Best Answer
Okay, I don't see that Ahlfors intended any particular method to be used (doesn't mean he didn't), so let's just use any method we can find, dominated convergence is the first that comes to mind.
First, for convenience, let's rotate it and consider
$$\int_{-\pi}^\pi \log\,\lvert 1 - e^{i\vartheta}\rvert \, d\vartheta,$$
so that we have the problematic point at $\vartheta = 0$.
A logarithmic singularity is integrable, since
$$ \int_\varepsilon^1 \log t\, dt = \left[t\log t -t\right]_\varepsilon^1 = -1 - \varepsilon\log\varepsilon + \varepsilon \to -1$$
for $\varepsilon \to 0$.
On the ray $r\cdot e^{i\vartheta}$ (for $\lvert \vartheta\rvert < \pi/2$), we consider
$$\delta(r) = \lvert 1- re^{i\vartheta}\rvert^2 = (1 - r\cos\vartheta)^2 + (r\sin\vartheta)^2 = 1+r^2 - 2r\cos\vartheta.$$
Since $\delta'(r) = 2(r - \cos\vartheta)$, $\delta(r)$ is minimised for $r = \cos\vartheta$, and $\delta(\cos\vartheta) = 1 - \cos^2\vartheta = \sin^2\vartheta$. That means that for any $0 < \varepsilon < \frac{\pi}{3}$ we have $\lvert \sin \vartheta\rvert \leqslant \lvert 1 - re^{i\vartheta}\rvert \leqslant 1$ for all $r\in [0,1]$ and $\lvert\vartheta\rvert \leqslant \varepsilon$, so the integrands $f_r(\vartheta) = \log\,\lvert 1 - re^{i\vartheta}\rvert$ are dominated by the integrable function $\log \frac{1}{\lvert\sin\vartheta\rvert}$ on $[-\varepsilon,\,\varepsilon]$. Outside that interval everything is bounded and convergence is uniform, so
$$\int_{-\pi}^\pi \log \,\lvert 1 - e^{i\vartheta}\rvert\,d\vartheta = \lim_{r \to 1} \int_{-\pi}^\pi \log\, \lvert 1 - re^{i\vartheta}\rvert\,d\vartheta = 0.$$
Another method that can be useful when you have a non-integrable limit, but such that the principal value of the integral exists (when you have a simple pole on the integration path, for example), is the removal of a small disk around the problematic point.
Now it would be good if I could draw a picture here, but since I can't, a description must do. Consider
$$B_\varepsilon := \mathbb{D}\setminus \overline{D_\varepsilon(1)} = \{z \in \mathbb{C} \colon \lvert z\rvert < 1, \lvert z-1\rvert > \varepsilon\}.$$
In a (simply connected) neighbourhood of $\overline{B}_\varepsilon$, the integrand is the real part of a holomorphic function ($\log (1-z)$), hence the (multiple of the) Cauchy integral
$$\int_{\partial B_\varepsilon} \frac{\log (1-z)}{z - z_0}\,dz$$
is $2\pi i \log (1-z_0)$ and vanishes for $z_0 = 0$.
The real part of that integral is then
$$\int_{\delta}^{2\pi-\delta} \log\, \lvert 1 - e^{i\vartheta}\rvert\, d\vartheta - \operatorname{Re} \int_{\partial B_\varepsilon \cap \mathbb{D}} \frac{\log (1-z)}{z}\,dz,$$
where $\delta = 2\arcsin(\varepsilon/2)$. The first integral is in the limit what we want, so we need to show that the second integral converges to $0$ for $\varepsilon \to 0$. The standard estimate does that,
$$\left\lvert \int_{\partial B_\varepsilon \cap \mathbb{D}} \frac{\log (1-z)}{z}\,dz\right\rvert \leqslant \pi\varepsilon \cdot \frac{\lvert\log \varepsilon\rvert + \pi/2}{1-\varepsilon}.$$