Prove that Cauchy transform diverges at 1 I found this problem on Lang Complex Analysis chapter VIII.
Le consider the function $g$ defined on the unit circle by:
$g(e^{ix})=
\begin{cases}
-\frac x{\pi\log(4/\pi)} & -\pi\leq x\leq 0\\
\frac 1{\log(4/x)} & 0<x<\pi
\end{cases}$
Then $g$ is continuous and we can take its Cauchy transform:
$$h(z)=\frac 1{2\pi i}\oint_{|\zeta|=1}\frac{g(\zeta)}{\zeta-z}d\zeta$$
for $z$ in the unit disc.
We have to prove that $h(z)\to \infty$ as $z\to 1$ on the real axis.
I tried to solve this problem by espliciting the integral for $h$, and I reduce it to show that the integral:
$$\int_0^\pi\frac{(1-\epsilon)\sin(x)}{\log(4/x)(2(1-\epsilon)(1-\cos(x))+\epsilon^2)}dx\to\infty$$
as $\epsilon\to 0^+$.

How can I prove that this integral diverges?

 A: $$\int_0^\pi\frac{\sin(x)}{\log(4/x)(2(1-\epsilon)(1-\cos(x))+\epsilon^2)}dx$$
The function inside the integrale is non-negative.
Since $\sin(x)\geq \frac x2$ for $0\leq x\leq 1$ we get
$$\int_0^\pi\frac{\sin(x)}{\log(4/x)(2(1-\epsilon)(1-\cos(x))+\epsilon^2)}dx\geq \int_0^1\frac{\sin(x)}{\log(4/x)(2(1-\epsilon)(1-\cos(x))+\epsilon^2)}dx\\ \geq\frac 12\int_0^1\frac x{\log(4/x)(2(1-\epsilon)(1-\cos(x))+\epsilon^2)}dx$$
We have $2(1-\epsilon)(1-\cos(x))\leq 2(1-\cos(x))\leq x^2$, from which:
$$\int_0^1\frac x{\log(4/x)(2(1-\epsilon)(1-\cos(x))+\epsilon^2)}dx\geq\int_0^1\frac x{\log(4/x)(x^2+\epsilon^2)}dx$$
Substitution $x=\sqrt t$ gives:
$$\int_0^1\frac x{\log(4/x)(x^2+\epsilon^2)}dx=\int_0^1\frac 1{-\log(t/16)(t+\epsilon^2)}dt$$
while with $u=t+\epsilon^2$ we get:
$$\int_0^1\frac 1{-\log(t/16)(t+\epsilon^2)}dt=\int_{\epsilon^2}^{1+\epsilon^2}\frac 1{-u\log((u-\epsilon^2)/16)}du\geq \int_{\epsilon}^1\frac 1{-u\log((u-\epsilon^2)/16)}du$$.
For $\epsilon\leq u\leq 1$ we have:
$$\frac{-\log((u-\epsilon^2)/16)}{-\log(u/16)}=\frac{-\log(u/16)-\log(1-\epsilon^2/u)}{-\log(u/16)}=1+\frac{-\log(1-\epsilon^2/u)}{-\log(u/16)}\\ \leq 1+\frac{-\log(1-\epsilon)}{-\log(1/16)}\leq 2$$
hence $0<-\log((u-\epsilon^2)/16)<-2\log(u/16)$, frmo which:
$$\int_{\epsilon}^1\frac 1{-u\log((u-\epsilon^2)/16)}du\geq\frac 12\int_{\epsilon}^1\frac 1{-u\log(u/16)}du\to +\infty$$
as $\epsilon\to 0^+$.
