复围道积分求解咨询：计算∫_γ z/(e^z-1)dz

Hey, let's break this down step by step to solve this integral—you're on the right track thinking about the Cauchy theorem with residues, so let's clarify the singularity details first.

Step 1: Analyze the Singularities of $f(z) = \frac{z}{e^z - 1}$ #

First, find all points where the function is not analytic:

• The denominator $e^z - 1 = 0$ when $z = 2\pi i n$ for any integer $n$ (since $e^{2\pi i n} = 1$ by Euler's formula).

• Removable Singularity at $z=0$: When $z \to 0$, use the Taylor expansion of $e^z$:
  $$e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots$$
  So $e^z - 1 = z\left(1 + \frac{z}{2} + \frac{z^2}{6} + \dots\right)$, and dividing by $z$ gives:
  $$f(z) = \frac{1}{1 + \frac{z}{2} + \frac{z^2}{6} + \dots}$$
  The limit as $z \to 0$ is 1, so $z=0$ is a removable singularity. Removable singularities have a residue of 0, so this point contributes nothing to the integral—we can ignore it entirely.

• First-Order Poles at $z=2\pi i n$ ($n \neq 0$): For non-zero integers $n$, $z=2\pi i n$ is a first-order pole. We confirm this by checking the derivative of the denominator: $h'(z) = e^z$, and $h'(2\pi i n) = e^{2\pi i n} = 1 \neq 0$. Since the denominator has a simple root here, the pole is first-order.

Step 2: Calculate Residues for the First-Order Poles #

For a first-order pole at $a$, if $f(z) = \frac{g(z)}{h(z)}$ where $g(a) \neq 0$, $h(a)=0$, and $h'(a) \neq 0$, the residue is given by:
$$\text{Res}(f, a) = \frac{g(a)}{h'(a)}$$
Here, $g(z) = z$ and $h(z) = e^z - 1$, so:
$$\text{Res}(f, 2\pi i n) = \frac{2\pi i n}{e^{2\pi i n}} = 2\pi i n$$
(Again, $e^{2\pi i n} = 1$ for integer $n$.)

Step 3: Apply the Cauchy Residue Theorem #

The theorem states that:
$$\int_\gamma f(z) dz = 2\pi i \times \sum \text{Res}(f, a)$$
where the sum includes all isolated singularities inside the contour $\gamma$.

Since $z=0$ contributes 0, we only sum the residues of the two poles inside $\gamma$. Let's say the two poles are $z=2\pi i k$ and $z=2\pi i m$ (for non-zero integers $k, m$):

• Sum of residues: $2\pi i k + 2\pi i m = 2\pi i (k + m)$
• Multiply by $2\pi i$:
  $$\int_\gamma f(z) dz = 2\pi i \times 2\pi i (k + m) = -4\pi^2 (k + m)$$

Example Case #

If the contour encloses $z=2\pi i$ and $z=-2\pi i$, the sum of residues is $2\pi i - 2\pi i = 0$, so the integral equals $0$. If it encloses $z=2\pi i$ and $z=4\pi i$, the integral becomes $2\pi i \times (2\pi i + 4\pi i) = -12\pi^2$.

内容的提问来源于stack exchange，提问作者RanSch