关于围道积分∫_{|z-i|=1} \frac{1}{z^2+1}dz的柯西定理应用正确性咨询

Hey Pascal, let's break down your work step by step to sort out what's right and where a small slip-up occurred!

First, your initial factoring is totally on point:
$$
\int_{|z-i|=1} \frac {1}{z^2+1}dz = \int_{|z-i|=1} \frac {1}{(z-i)(z+i)}dz
$$
This is exactly the right first move for applying Cauchy's theorems here.

But here's the tiny mistake: you wrote the integrand as $\frac {\frac {1}{z-1}}{(z+i)}$, which is a typo. You should have rearranged it to $\frac {\frac{1}{z+i}}{z-i}$ instead. The key idea is to shape the integrand into the form $\frac{f(z)}{z-a}$, where $a$ is the singularity inside the contour $|z-i|=1$ (that's $a=i$ here), and $f(z)$ is analytic everywhere inside the contour.

Let's confirm $f(z)=\frac{1}{z+i}$ is analytic inside the contour: the point $z=-i$ is 2 units away from $i$ (since $|i - (-i)|=2$), which is farther than the contour's radius of 1. So $f(z)$ has no singularities inside the contour, which means we can safely use the Cauchy Integral Formula:
$$
\int_{C} \frac{f(z)}{z-a} dz = 2\pi i \cdot f(a)
$$

Plugging in $a=i$ and $f(z)=\frac{1}{z+i}$:
$$
\int_{|z-i|=1} \frac{\frac{1}{z+i}}{z-i}dz = 2\pi i \cdot f(i) = 2\pi i \cdot \frac{1}{i+i} = 2\pi i \cdot \frac{1}{2i} = \pi
$$

So your final conclusion that the integral equals $\pi$ is correct! The earlier $-\pi$ result came from that typo in the fraction rearrangement. Your overall approach using Cauchy's theorem (specifically the integral formula) is completely sound — you just had a tiny writing error in the middle step.

No worries at all, glad to help clear that up!

备注：内容来源于stack exchange，提问作者Pascal