3.20. Calculate

$$
\int_{1}^{i} \frac{\ln ^{2} z}{z} d z
$$
Solution. Let's take some region $D$ that does not contain 0. In this region, as it is not difficult to verify, the function

$$
f(z)=\frac{\ln ^{2} z}{z}
$$

is analytic, and therefore, by the corollary of Cauchy's theorem,

$$
\begin{gathered}
\int_{1}^{i} \frac{\ln ^{2} z}{z} d z=\int_{1}^{i} \ln ^{2} z d \ln z=\left.\frac{\ln ^{3} z}{3}\right|_{1} ^{i}=\frac{\ln ^{3} i-\ln ^{3} 1}{3}= \\
=\frac{\ln ^{3} i}{3}=\frac{(\ln |i|+i \arg i)^{3}}{3}=\frac{1}{3}\left(i \frac{\pi}{2}\right)^{3}=\frac{-i \pi^{3}}{24}
\end{gathered}
$$

### 3.12 .

CAUCHY INTEGRAL FORMULA

### 3.12.1.  CAUCHY INTEGRAL FORMULA

Let $f(z)$ be a single-valued analytic function in a simply connected domain $D$ and $L \subset D$ be an arbitrary piecewise continuous closed curve bounding a region $G \subset D$. Then for any point $z_{0} \in G$, the Cauchy integral formula holds:

$$
f\left(z_{0}\right)=\frac{1}{2 \pi i} \int_{L} \frac{f(z)}{z-z_{0}} d z
$$

The Cauchy formula allows us to determine the value of the analytic function $f(z)$ at any point inside a certain region of analyticity based on its values on a closed contour bounding this region.

3.12.2.

FORMULA FOR DERIVATIVES

For any point $z_{0} \in G \subset D$, the derivative $f^{\prime}\left(z_{0}\right)$ can be found by differentiating the right-hand side of (3.43) under the integral sign, leading to the following expression for $f^{\prime}\left(z_{0}\right):$

$$
f^{\prime}\left(z_{0}\right)=\frac{1}{2 \pi i} \int \frac{f(z)}{\left(z-z_{0}\right)^{2}} d z
$$

By applying repeated differentiation to (3.46), it can be shown by induction that the formula for the $n$-th order derivative of the analytic function $f(z)$ is:

$$
f^{(n)}\left(z_{0}\right)=\frac{n!}{2 \pi i} \int \frac{f(z)}{\left(z-z_{0}\right)^{n+1}} d z
$$

## Examples.