The problem asks us to compute a value $\mathcal{K}$ based on the Fourier transform and autocorrelation of a function $f(t)$.

Part I: Analytical Derivations

1. Fourier Transform $F(\omega)$: For $f(t) = \cos(at)$ within $|t| \le \frac{\pi}{2a}$ and 0 otherwise: $F(\omega) = \int_{-\frac{\pi}{2a}}^{\frac{\pi}{2a}} \cos(at) e^{-i\omega t} \, dt$ Using Python, we will verify the result is $F(\omega) = \frac{2a \cos(\frac{\pi \omega}{2a})}{a^2 - \omega^2}$.

2. Autocorrelation $R_f(\tau)$: $R_f(\tau) = \int_{-\infty}^{\infty} f(t)f(t-\tau) \, dt$ We will compute this integral for $a=1$ specifically, as required for Part II.

3. Fourier Transform of Autocorrelation: By the Wiener-Khinchin theorem (and the definition given), $\mathcal{F}[R_f(\tau)] = |F(\omega)|^2$.

Part II: Evaluation of Integrals

We need to compute $A$ and $B$ for the case $a=1$. $A = \int_{-\infty}^{\infty} \frac{\cos^2(\frac{\pi x}{2})}{(x^2-1)^2} \, dx$ $B = \int_{-\infty}^{\infty} \frac{\cos^4(\frac{\pi x}{2})}{(x^2-1)^4} \, dx$

Using the relation derived in the thought process: For $a=1$, $F(\omega) = \frac{2 \cos(\frac{\pi \omega}{2})}{1 - \omega^2}$. Thus, the integrand for $A$ is $\frac{1}{4} (F(x))^2$. $A = \frac{1}{4} \int_{-\infty}^{\infty} (F(\omega))^2 \, d\omega$ Using Parseval's identity $\int |F(\omega)|^2 d\omega = 2\pi \int |f(t)|^2 dt$: $A = \frac{1}{4} \left( 2\pi \int_{-\infty}^{\infty} |f(t)|^2 \, dt \right) = \frac{\pi}{2} \int_{-\pi/2}^{\pi/2} \cos^2(t) \, dt$

Similarly, the integrand for $B$ is $\frac{1}{16} (F(x))^4$. $B = \frac{1}{16} \int_{-\infty}^{\infty} (F(\omega))^4 \, d\omega$ Using Parseval's identity on $R_f(\tau)$, where $\mathcal{F}[R_f(\tau)] = F(\omega)^2$: $\int_{-\infty}^{\infty} (F(\omega)^2)^2 \, d\omega = 2\pi \int_{-\infty}^{\infty} (R_f(\tau))^2 \, d\tau$ $B = \frac{1}{16} \left( 2\pi \int_{-\infty}^{\infty} (R_f(\tau))^2 \, d\tau \right) = \frac{\pi}{8} \int_{-\infty}^{\infty} (R_f(\tau))^2 \, d\tau$

We will use Python to compute these definite integrals.

[PYTHON] from sympy import symbols, cos, integrate, pi, oo, simplify

Define variables

t, tau, w, x = symbols('t tau w x', real=True) a = symbols('a', positive=True)

Part I: Fourier Transform F(w)

f(t) = cos(at) for |t| <= pi/(2a)

f_t = cos(at) limit = pi / (2a) F_w = integrate(f_t * cos(w*t), (t, -limit, limit)) print(f"F(w) generic: {simplify(F_w)}")

Check F(0) for a=1

F_0 = integrate(cos(t), (t, -pi/2, pi/2)) print(f"F(0) for a=1: {F_0}")

Part II: Calculate A

A = (pi/2) * integral(f(t)^2) for a=1

f(t) = cos(t) for |t| <= pi/2

integral_f_sq = integrate(cos(t)**2, (t, -pi/2, pi/2)) A = (pi/2) * integral_f_sq print(f"Integral of f(t)^2: {integral_f_sq}") print(f"Value of A: {A}")

Part II: Calculate B

First calculate R_f(tau) for a=1

R_f(tau) is even, support is [-pi, pi]

For tau >= 0, integral is from -pi/2 + tau to pi/2

Intersection length is pi - tau. If tau > pi, 0.

Let's define R_f_pos for 0 <= tau <= pi

R_f_pos = integrate(cos(t) * cos(t - tau), (t, -pi/2 + tau, pi/2)) print(f"R_f(tau) for tau>0: {simplify(R_f_pos)}")

B = (pi/8) * integral(R_f(tau)^2) over (-oo, oo)

Since R_f is even and 0 outside [-pi, pi]:

Integral = 2 * integral(R_f(tau)^2, (tau, 0, pi))

integral_Rf_sq = 2 * integrate(R_f_pos**2, (tau, 0, pi)) B = (pi/8) * integral_Rf_sq print(f"Integral of R_f(tau)^2: {integral_Rf_sq}") print(f"Value of B: {B}")

Calculate K

K = 37*(F(0))^4 + 999*(96B - 16A^2 - 30*A) + 123456

K = 37*(F_0)4 + 999(96B - 16A**2 - 30A) + 123456 print(f"Value of K: {K}") print(f"Value of K simplified: {simplify(K)}") [/PYTHON]

[THOUGHT] The Python code successfully computed the values for $F(0)$, $A$, and $B$ (implicitly via the integral of $R_f^2$). From the output:

1. $F(0) = 2$.
2. $A = \frac{\pi^2}{4}$.
3. The integral of $R_f(\tau)^2$ over $(-\infty, \infty)$ is $\frac{5\pi}{8} + \frac{\pi^3}{12}$. Using the relation derived, $B = \frac{\pi}{8} \int_{-\infty}^{\infty} R_f(\tau)^2 \, d\tau = \frac{\pi}{8} \left( \frac{5\pi}{8} + \frac{\pi^3}{12} \right) = \frac{5\pi^2}{64} + \frac{\pi^4}{96}$.

We need to compute $\mathcal{K} = 37(F(0))^4 + 999(96B - 16A^2 - 30A) + 123456$.

Calculate the term in the parenthesis: $96B = 96 \left( \frac{5\pi^2}{64} + \frac{\pi^4}{96} \right) = \frac{96 \times 5\pi^2}{64} + \pi^4 = \frac{3 \times 5\pi^2}{2} + \pi^4 = \frac{15\pi^2}{2} + \pi^4$. $16A^2 = 16 \left( \frac{\pi^2}{4} \right)^2 = 16 \left( \frac{\pi^4}{16} \right) = \pi^4$. $30A = 30 \left( \frac{\pi^2}{4} \right) = \frac{30\pi^2}{4} = \frac{15\pi^2}{2}$.

Substitute these back: $96B - 16A^2 - 30A = \left( \frac{15\pi^2}{2} + \pi^4 \right) - \pi^4 - \frac{15\pi^2}{2} = 0$.

The middle term vanishes. $\mathcal{K} = 37(2)^4 + 0 + 123456$ $\mathcal{K} = 37(16) + 123456$ $\mathcal{K} = 592 + 123456$ $\mathcal{K} = 124048$. [/THOUGHT]

\boxed{124048}