[THOUGHT] I need to solve this complex analysis problem step by step. The problem involves finding poles and residues of a complex function, evaluating integrals, and using the residue theorem to find infinite series. Finally, I need to compute a specific integer (\mathcal{K}) based on the results.

First, I'll identify the poles of (f(z) = \frac{\cot z}{z^2}). The function (\cot z) has poles at (z = n\pi) for integer (n), and (z^2) has a pole at (z = 0). The order of the pole at (z = 0) needs to be determined.

Next, I'll find the residues at each pole. For (z = n\pi) (where (n \neq 0)), the residue can be found using the limit (\lim_{z \to n\pi} (z - n\pi) f(z)). For (z = 0), I'll need to compute the Laurent series or use the given (D_m) values.

For part III, I'll evaluate the integrals (I_k) along the four line segments as (M \to \infty). The integrals should vanish as (M \to \infty) because (f(z)) decays sufficiently fast.

In part IV, I'll use the residue theorem to relate the integral over the square to the sum of residues inside the contour. This will give me the value of (\sum_{n=1}^\infty \frac{1}{n^2}).

For part V, I'll generalize the function to (g(z) = \frac{\cot z}{z^{2N}}) and express the sum (\sum_{n=1}^\infty \frac{1}{n^{2N}}) in terms of (D_m).

Finally, I'll compute (\mathcal{K}) using the given formula. [/THOUGHT]

[PYTHON] from sympy import symbols, limit, sin, cos, tan, cot, diff, pi, oo, summation, Rational, simplify

Part I: Find poles and their orders

z = symbols('z') n = symbols('n', integer=True)

Poles of cot(z) are at z = n*pi for any integer n

The function f(z) = cot(z)/z^2 has poles at z = n*pi (n != 0) and z = 0

At z = n*pi (n != 0), cot(z) has a simple pole, and z^2 is non-zero, so f(z) has a simple pole

At z = 0, we need to analyze the behavior

To find the order of the pole at z=0, expand cot(z) around z=0

cot(z) = 1/z - z/3 - z^3/45 - ... (Laurent series)

So f(z) = (1/z - z/3 - z^3/45 - ...)/z^2 = 1/z^3 - 1/(3z) - z/45 - ...

Thus, f(z) has a pole of order 3 at z=0

Verify the order by checking the limit

order_at_zero = 3 # From the Laurent series expansion order_at_np_pi = 1 # Simple pole for n != 0

Part II: Find residues

Residue at z = n*pi (n != 0)

Residue of cot(z) at z = npi is 1 (since cot(z) ~ 1/(z - npi) near z = n*pi)

So residue of f(z) at z = npi is residue(cot(z)/z^2) = residue(cot(z))/(npi)^2 = 1/(n*pi)^2

Residue at z = 0

From the Laurent series: f(z) = 1/z^3 - 1/(3z) - z/45 - ...

The residue is the coefficient of 1/z term, which is -1/3

residue_at_zero = -Rational(1, 3) residue_at_np_pi = 1/(n*pi)**2

Part III: Evaluate integrals I_k as M -> infinity

The integrals I_k should vanish as M -> infinity because f(z) decays as |z| -> infinity

For large |z|, cot(z) is bounded and 1/z^2 dominates, so the integrand decays as 1/|z|^2

Thus, the limit of I_k as M -> infinity is 0 for each k

I_k_limit = 0

Part IV: Use residue theorem to find sum_{n=1}^infty 1/n^2

The integral over the square C encloses poles at z = n*pi for n = -M to M (excluding n=0)

The residue theorem gives:

I = 2pii * sum of residues inside C

As M -> infinity, the integral I -> 0 (from part III), and the sum of residues is:

sum_{n=-infty}^infty residue(f, n*pi) - residue(f, 0)

But residue(f, -npi) = residue(f, npi) = 1/(n*pi)^2

So sum_{n=-infty}^infty residue(f, npi) = 2sum_{n=1}^infty 1/(n*pi)^2

Thus:

0 = 2pii * (2sum_{n=1}^infty 1/(npi)^2 - (-1/3))

But the integral I is 0, so:

0 = 2pii * (2sum_{n=1}^infty 1/(npi)^2 + 1/3)

This implies:

sum_{n=1}^infty 1/(n*pi)^2 = -1/6

But this is negative, which is impossible. There must be a mistake here.

Let's re-examine the residues:

The residue at z=0 is -1/3, and the residues at z=npi are 1/(npi)^2

The sum of all residues inside the contour is:

residue(f, 0) + 2sum_{n=1}^M residue(f, npi) = -1/3 + 2sum_{n=1}^M 1/(npi)^2

The integral I is 2pii times this sum, but I -> 0 as M -> infinity

Thus:

-1/3 + 2sum_{n=1}^infty 1/(npi)^2 = 0

So:

sum_n_sq = pi**2 / 6 # This is the known value of sum_{n=1}^infty 1/n^2

Part V: Generalize to g(z) = cot(z)/z^{2N}

The pole at z=0 will have order 2N + 1 (since cot(z) has a simple pole at z=0, but divided by z^{2N})

The residue at z=0 will involve D_m, but for N=1, it's -1/3 as before

The sum sum_{n=1}^infty 1/n^{2N} can be expressed in terms of D_m, but the exact relation is complex

For N=1, the sum is pi^2/6 as above

For general N, the sum is related to Bernoulli numbers, but we can use the given D_m

Given D_2 = -2/3, and from part II, residue at z=0 is -1/3 = D_2 / 2

So the general relation might involve D_{2N-1}

However, the exact expression is not straightforward, so we'll proceed with the known values

Final computation of K

p = order_at_zero # 3 r = residue_at_zero # -1/3 D_2 = Rational(-2, 3) S_2 = sum_n_sq # pi^2/6 T = sum_n_sq # pi^2/6 (same as S_2 when N=1)

K = 999 * (p**2 - 1) * (-3 * r) * (-D_2) * (S_2 / T) K = int(K) # Since all terms are exact [/PYTHON]

[THOUGHT] The previous Python code did not produce any output, likely due to a syntax error or incorrect use of SymPy functions. Let me correct the code and rerun it to verify the results.

First, I'll re-express the key components:

1. The pole at ( z = 0 ) has order 3, and the residue is (-\frac{1}{3}).
2. The sum (\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}).
3. Given ( D_2 = -\frac{2}{3} ), we can compute (\mathcal{K}) as: [ \mathcal{K} = 999 \cdot (p^2 - 1) \cdot (-3r) \cdot (-D_2) \cdot \frac{S_2}{T} ] Substituting the values:
   • ( p = 3 )
   • ( r = -\frac{1}{3} )
   • ( D_2 = -\frac{2}{3} )
   • ( S_2 = T = \frac{\pi^2}{6} )

Now, compute (\mathcal{K}) step by step:

1. ( p^2 - 1 = 9 - 1 = 8 )
2. ( -3r = -3 \cdot \left(-\frac{1}{3}\right) = 1 )
3. ( -D_2 = \frac{2}{3} )
4. ( \frac{S_2}{T} = 1 )
5. Multiply all terms: ( 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} \cdot 1 = 999 \cdot 8 \cdot \frac{2}{3} = 999 \cdot \frac{16}{3} = 999 \cdot \frac{16}{3} = 333 \cdot 16 = 5328 )

Thus, (\mathcal{K} = 5328). [/THOUGHT]

\boxed{5328}