方形围道复分析问题求证：正弦模估计与积分极限证明

Let's break this problem down step by step, starting with part (a) since it lays the groundwork for proving part (b).

Part (a): Bounding $|\sin(z)|$ on Contour $S_N$ #

First, let's recap the contour $S_N$: it's a square centered at the origin with side length $(2N+1)\pi$, vertices at $\pi(N+1/2)\pm i\pi(N+1/2)$ and $-\pi(N+1/2)\pm i\pi(N+1/2)$. We'll analyze vertical and horizontal edges separately.

Vertical Edges #

The vertical edges are:

• Right edge: $z = \pi(N+1/2) + iy$, where $y \in [-\pi(N+1/2), \pi(N+1/2)]$
• Left edge: $z = -\pi(N+1/2) + iy$, where $y \in [-\pi(N+1/2), \pi(N+1/2)]$

Using the identity for $\sin(z)$ when $z = x + iy$:
$$\sin(x+iy) = \sin x \cosh y + i\cos x \sinh y$$

For the right edge, $x = \pi(N+1/2)$, so $\cos x = \cos\left(\pi N + \pi/2\right) = 0$, and $\sin x = \sin\left(\pi N + \pi/2\right) = (-1)^N$. The modulus simplifies to:
$$|\sin(z)| = \sqrt{(\sin x \cosh y)^2 + (\cos x \sinh y)^2} = |\sin x| \cdot |\cosh y| = |\cosh y|$$

Since $\cosh y = \frac{e^y + e^{-y}}{2} \ge 1$ for all real $y$ (by AM-GM inequality), we get $|\sin(z)| \ge 1$ on the right vertical edge. The left vertical edge is symmetric: $\sin(-x+iy) = -\sin x \cosh y + i\cos x \sinh y$, so its modulus is also $|\cosh y| \ge 1$.

Horizontal Edges #

The horizontal edges are:

• Upper edge: $z = x + i\pi(N+1/2)$, where $x \in [-\pi(N+1/2), \pi(N+1/2)]$
• Lower edge: $z = x - i\pi(N+1/2)$, where $x \in [-\pi(N+1/2), \pi(N+1/2)]$

Let $a = \pi(N+1/2)$ for simplicity. Using the $\sin(z)$ identity again:
$$\sin(x+ia) = \sin x \cosh a + i\cos x \sinh a$$

Calculate the modulus squared:
$$|\sin(z)|^2 = (\sin x \cosh a)^2 + (\cos x \sinh a)^2$$
Substitute $\cosh^2 a = \sinh^2 a + 1$ to simplify:
$$= \sin^2 x (\sinh^2 a + 1) + \cos^2 x \sinh^2 a = \sinh^2 a + \sin^2 x$$

Since $\sin^2 x \ge 0$, we have $|\sin(z)| \ge \sinh a$. Now, $a = \pi(N+1/2) \ge \pi/2$ (for $N \ge 1$), and $\sinh(t)$ is increasing for $t > 0$. Thus:
$$\sinh a \ge \sinh(\pi/2)$$
So $|\sin(z)| \ge \sinh(\pi/2)$ on the upper horizontal edge. The lower edge is symmetric: $\sin(x-ia)$ has the same modulus squared, leading to the same lower bound.

Part (b): Proving the Integral Limit is Zero #

We'll use the ML Inequality (Estimation Lemma) here: for a contour $\Gamma$ of length $L$, if $|f(z)| \le M$ for all $z \in \Gamma$, then $\left|\int_\Gamma f(z)dz\right| \le M \cdot L$.

First, compute the total length of $S_N$: each side is $(2N+1)\pi$, so perimeter $L = 4(2N+1)\pi$.

Next, find an upper bound for $|f(z)| = \left|\frac{1}{z^2 \sin(z)}\right|$ on each edge:

Bounding $|f(z)|$ on Vertical Edges #

On vertical edges, $|z| = \sqrt{[\pi(N+1/2)]^2 + y^2} \ge \pi(N+1/2)$ (since $y^2 \ge 0$). Thus $|z^2| \ge \pi^2(N+1/2)2$. From part (a), $|\sin(z)| \ge 1$, so:
$$|f(z)| \le \frac{1}{\pi^2(N+1/2)2 \cdot 1} = \frac{1}{\pi^2(N+1/2)2}$$

Bounding $|f(z)|$ on Horizontal Edges #

On horizontal edges, $|z| \ge \pi(N+1/2)$ (same reasoning as vertical edges), so $|z^2| \ge \pi^2(N+1/2)2$. From part (a), $|\sin(z)| \ge \sinh(\pi/2)$ (a positive constant, let's call it $C = \sinh(\pi/2)$). Thus:
$$|f(z)| \le \frac{1}{\pi^2(N+1/2)2 \cdot C}$$

Applying the ML Inequality #

Split the integral into four edge contributions:
$$\left|\int_{S_N} f(z)dz\right| \le \sum_{\text{edges}} \left|\int_{\text{edge}} f(z)dz\right|$$

Each vertical edge has length $(2N+1)\pi$, each horizontal edge also has length $(2N+1)\pi$. Substitute the bounds:
$$\le 2 \cdot (2N+1)\pi \cdot \frac{1}{\pi^2(N+1/2)2} + 2 \cdot (2N+1)\pi \cdot \frac{1}{\pi^2(N+1/2)2 C}$$

Notice $2N+1 = 2(N+1/2)$, so substitute that in to simplify:
$$= 2 \cdot 2(N+1/2)\pi \cdot \frac{1}{\pi^2(N+1/2)2} \left(1 + \frac{1}{C}\right) = \frac{4}{\pi(N+1/2)} \left(1 + \frac{1}{C}\right)$$

As $N \rightarrow \infty$, $(N+1/2) \rightarrow \infty$, so the entire expression tends to 0. Therefore:
$$\lim_{N\rightarrow\infty}\left|\int_{S_N} \frac{1}{z^{2} \sin(z)}dz\right| = 0$$

内容的提问来源于stack exchange，提问作者thrillho