Problem Analysis #2

[THOUGHT] First, I need to identify the poles of $f(z) = \frac{\cot z}{z^2}$.

The cotangent function $\cot z = \frac{\cos z}{\sin z}$ has poles where $\sin z = 0$. These occur at $z = n\pi$ for integers $n$, and each of these is a simple pole of $\cot z$.

Therefore, $f(z) = \frac{\cot z}{z^2}$ has poles at $z = n\pi$ for all integers $n$, including $n = 0$.

Order of each pole:

1. At $z = 0$: We need to determine the order. Since $\cot z$ behaves like $\frac{1}{z}$ near $z = 0$, $\frac{\cot z}{z^2}$ behaves like $\frac{1}{z^3}$. So, the pole at $z = 0$ is of order 3.

2. At $z = n\pi$ for $n \neq 0$: Since $\cot z$ has a simple pole at these points, dividing by $z^2$ results in a pole of order 3 (one from $\cot z$ and two from $z^2$).

Residues at each pole:

1. At $z = 0$: To find the residue of $f(z)$ at $z = 0$, we use the definition of the residue for a pole of order 3:

$$\text{Res}(f(z), 0) = \frac{1}{2!} \lim_{z \to 0} \frac{d^2}{dz^2} \left( z^3 f(z) \right) = \frac{1}{2} \lim_{z \to 0} \frac{d^2}{dz^2} (z \cot z).$$

We are given $D_2 = -\frac{2}{3}$, so:

$$\text{Res}(f(z), 0) = \frac{1}{2} \left( -\frac{2}{3} \right) = -\frac{1}{3}.$$

2. At $z = n\pi$ for $n \neq 0$: We use the fact that $\cot z$ has residues of $(-1)^n$ at these points. The residue of $\frac{\cot z}{z^2}$ at $z = n\pi$ is:

$$\text{Res}(f(z), n\pi) = \lim_{z \to n\pi} \frac{z - n\pi}{(\cot z)(z - n\pi)^2} \cdot (z - n\pi)^3 = \lim_{z \to n\pi} \frac{z - n\pi}{\sin z} \cdot \frac{\cos z}{(z - n\pi)^2} = (-1)^n \cos(n\pi) \cdot \frac{1}{(n\pi)^2}.$$

Since $\cos(n\pi) = (-1)^n$, we get:

$$\text{Res}(f(z), n\pi) = (-1)^n \cdot (-1)^n \cdot \frac{1}{(n\pi)^2} = \frac{1}{(n\pi)^2}.$$

Integral along the contour $C_k$: For each $k = 1, 2, 3, 4$, the integral $I_k = \int_{C_k} f(z)\,dz$ along $C_k$ as $M \to \infty$:

1. $C_1$ and $C_3$: These are vertical lines. As $M \to \infty$, $|z| \to \infty$, and $\cot z$ is bounded. Therefore, $|f(z)| \to 0$, making each integral vanish.
2. $C_2$ and $C_4$: These are horizontal lines. As $M \to \infty$, the integrals also vanish because $\cot z$ oscillates between bounded values, and the length of the segment increases while the function's value decreases.

Thus, $\lim_{M \to \infty} I_k = 0$ for $k = 1, 2, 3, 4$.

Residue theorem application: The residue theorem states:

$$\oint_C f(z)\,dz = 2\pi i \sum \text{Residues inside } C.$$

As $M \to \infty$, the only poles inside $C$ are at $z = n\pi$ for $n = -M, -M+1, \ldots, M$.

The residues sum up as:

$$\sum_{n=-M, n\neq 0}^{M} \frac{1}{(n\pi)^2} + \text{Res}(f(z), 0) = \sum_{n=1}^{M} \frac{1}{(n\pi)^2} + \sum_{n=1}^{M} \frac{1}{(n\pi)^2} - \frac{1}{3} = 2 \sum_{n=1}^{M} \frac{1}{(n\pi)^2} - \frac{1}{3}.$$

Thus,

$$\oint_C f(z)\,dz = 2\pi i \left( 2 \sum_{n=1}^{M} \frac{1}{(n\pi)^2} - \frac{1}{3} \right).$$

Since $\oint_C f(z)\,dz \to 0$ as $M \to \infty$, we have:

$$0 = 2\pi i \left( 2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^2} - \frac{1}{3} \right).$$

Solving for $\sum_{n=1}^{\infty} \frac{1}{n^2}$, we get:

$$2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^2} = \frac{1}{3} \implies \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.$$

Generalizing for $g(z) = \frac{\cot z}{z^{2N}}$: The poles and residues for $g(z)$ follow a similar pattern. The residue at $z = n\pi$ for $n \neq 0$ becomes:

$$\text{Res}(g(z), n\pi) = \frac{1}{(n\pi)^{2N}}.$$

The residue at $z = 0$ involves higher-order derivatives and is more complex, but the key idea is that the residue theorem will yield:

$$\oint_C g(z)\,dz = 2\pi i \left( \text{Res}(g(z), 0) + 2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^{2N}} \right).$$

Since $\oint_C g(z)\,dz \to 0$ as $M \to \infty$, we solve for the series:

$$0 = 2\pi i \left( \text{Res}(g(z), 0) + 2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^{2N}} \right).$$

This implies:

$$\text{Res}(g(z), 0) = -2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^{2N}} \implies \sum_{n=1}^{\infty} \frac{1}{n^{2N}} = -\frac{\pi^{2N}}{2} \text{Res}(g(z), 0).$$

For $N = 1$, we have:

$$T = \sum_{n=1}^{\infty} \frac{1}{n^2} = -\frac{\pi^2}{2} \left( -\frac{1}{3} \right) = \frac{\pi^2}{6}.$$

Final Calculation: From I, the order $p = 3$. From II, the residue $r = -\frac{1}{3}$. From IV, $S_2 = \frac{\pi^2}{6}$. From the expression in V, $T = \frac{\pi^2}{6}$.

Using $D_2 = -\frac{2}{3}$, we compute:

$$\mathcal{K} = 999 \times (p^2 - 1) \times (-3r) \times (-D_2) \times \frac{S_2}{T}.$$

Since $S_2 = T$, the ratio $\frac{S_2}{T} = 1$, so:

$$\mathcal{K} = 999 \times (3^2 - 1) \times (-3 \times -\frac{1}{3}) \times (-\frac{2}{3}) \times 1 = 999 \times 8 \times 1 \times (-\frac{2}{3}) = 999 \times 8 \times -\frac{2}{3} = 999 \times -\frac{16}{3} = -3332.$$

The integer value of $\mathcal{K}$ is:

$$\boxed{-3332}$$