[THOUGHT] Let me work through this problem step by step.

Part I:

For $f(t) = \cos(at)$ when $|t| \le \frac{\pi}{2a}$ and $0$ otherwise.

(1) Fourier Transform $F(\omega)$:

$F(\omega) = \int_{-\pi/(2a)}^{\pi/(2a)} \cos(at) e^{-i\omega t} dt$

Using $\cos(at) = \frac{e^{iat} + e^{-iat}}{2}$:

$F(\omega) = \frac{1}{2} \int_{-\pi/(2a)}^{\pi/(2a)} (e^{i(a-\omega)t} + e^{-i(a+\omega)t}) dt$

$= \frac{1}{2} \left[ \frac{e^{i(a-\omega)t}}{i(a-\omega)} + \frac{e^{-i(a+\omega)t}}{-i(a+\omega)} \right]_{-\pi/(2a)}^{\pi/(2a)}$

$= \frac{\sin((a-\omega)\pi/(2a))}{a-\omega} + \frac{\sin((a+\omega)\pi/(2a))}{a+\omega}$

Let me simplify: $\sin((a-\omega)\frac{\pi}{2a}) = \sin(\frac{\pi}{2} - \frac{\omega\pi}{2a}) = \cos(\frac{\omega\pi}{2a})$

Similarly: $\sin((a+\omega)\frac{\pi}{2a}) = \sin(\frac{\pi}{2} + \frac{\omega\pi}{2a}) = \cos(\frac{\omega\pi}{2a})$

So $F(\omega) = \cos(\frac{\omega\pi}{2a}) \left( \frac{1}{a-\omega} + \frac{1}{a+\omega} \right) = \cos(\frac{\omega\pi}{2a}) \cdot \frac{2a}{a^2 - \omega^2}$

$F(\omega) = \frac{2a \cos(\frac{\omega\pi}{2a})}{a^2 - \omega^2}$

For $a=1$: $F(\omega) = \frac{2\cos(\frac{\omega\pi}{2})}{1 - \omega^2}$

$F(0) = 2$

(2) Autocorrelation $R_f(\tau)$:

This is the convolution of $f(t)$ with $f(-t)$. Since $f(t)$ is even, $R_f(\tau) = (f * f)(\tau)$.

(3) Fourier Transform of $R_f(\tau)$:

$\mathcal{F}[R_f(\tau)] = |F(\omega)|^2$

Part II:

For $a=1$: $|F(\omega)|^2 = \frac{4\cos^2(\frac{\omega\pi}{2})}{(1-\omega^2)^2}$

Using Parseval's theorem: $\int_{-\infty}^{\infty} |F(\omega)|^2 d\omega = 2\pi \int_{-\infty}^{\infty} |f(t)|^2 dt$

Let me compute the integrals using Python. [/THOUGHT] [PYTHON] import sympy as sp from sympy import symbols, cos, sin, pi, integrate, oo, Rational, simplify, sqrt

Part I with a=1

a = 1 t, tau, omega, x = symbols('t tau omega x', real=True)

(1) F(omega) for a=1

F(omega) = 2cos(omegapi/2) / (1 - omega^2)

F(0) = 2

F_0 = 2 print(f"F(0) = {F_0}")

Part II(1): Integral of cos^2(pi*x/2) / (x^2-1)^2 dx from -inf to inf

This is related to |F(omega)|^2 / 4 with omega = x

Using Parseval: integral |F(omega)|^2 d_omega = 2*pi * integral |f(t)|^2 dt

|f(t)|^2 = cos^2(t) for |t| <= pi/2, 0 otherwise

integral of cos^2(t) from -pi/2 to pi/2

int_f_squared = integrate(cos(t)**2, (t, -pi/2, pi/2)) print(f"Integral of |f(t)|^2 = {int_f_squared}")

So integral |F(omega)|^2 d_omega = 2*pi * (pi/2) = pi^2

int_F_squared = 2 * pi * int_f_squared print(f"Integral of |F(omega)|^2 = {int_F_squared}")

|F(omega)|^2 = 4cos^2(omegapi/2) / (1-omega^2)^2

So integral of 4cos^2(omegapi/2) / (omega^2-1)^2 d_omega = pi^2

Note: (1-omega^2)^2 = (omega^2-1)^2

Therefore: integral of cos^2(pi*x/2) / (x^2-1)^2 dx = pi^2 / 4

A = pi**2 / 4 print(f"A = {A}")

Part II(2): Integral of cos^4(pi*x/2) / (x^2-1)^4 dx

This relates to |F(omega)|^4 / 16

We need integral of |F(omega)|^4 d_omega

|F(omega)|^4 = |F(omega)|^2 * |F(omega)|^2

By Parseval on R_f: integral |F[R_f]|^2 d_omega = 2*pi * integral |R_f(tau)|^2 d_tau

And F[R_f] = |F(omega)|^2

So integral |F(omega)|^4 d_omega = 2*pi * integral |R_f(tau)|^2 d_tau

R_f(tau) = integral f(t) f(t-tau) dt

For f(t) = cos(t) on [-pi/2, pi/2], this is the convolution

Let me compute R_f(tau) directly

The support of R_f is [-pi, pi] (since support of f is [-pi/2, pi/2])

For |tau| <= pi, R_f(tau) = integral of cos(t)*cos(t-tau) dt over the overlap region

Actually, let me compute this more carefully

f(t) is nonzero for t in [-pi/2, pi/2]

f(t-tau) is nonzero for t-tau in [-pi/2, pi/2], i.e., t in [tau-pi/2, tau+pi/2]

For tau >= 0:

Overlap is [max(-pi/2, tau-pi/2), min(pi/2, tau+pi/2)]

For 0 <= tau <= pi: overlap is [tau-pi/2, pi/2] if tau <= pi/2, else [tau-pi/2, pi/2] ...

Let me just compute numerically first

import numpy as np from scipy import integrate as sci_integrate

def f_func(t): return np.cos(t) * (np.abs(t) <= np.pi/2)

def R_f_func(tau): def integrand(t): return f_func(t) * f_func(t - tau) result, _ = sci_integrate.quad(integrand, -np.pi, np.pi) return result

Compute integral of R_f(tau)^2

def R_f_squared(tau): return R_f_func(tau)**2

int_R_f_squared, _ = sci_integrate.quad(R_f_squared, -np.pi, np.pi) print(f"Integral of R_f(tau)^2 = {int_R_f_squared}")

integral |F(omega)|^4 d_omega = 2*pi * int_R_f_squared

int_F_fourth = 2 * np.pi * int_R_f_squared print(f"Integral of |F(omega)|^4 = {int_F_fourth}")

|F(omega)|^4 = 16 * cos^4(omega*pi/2) / (omega^2-1)^4

So integral of cos^4(pi*x/2) / (x^2-1)^4 dx = int_F_fourth / 16

B_numerical = int_F_fourth / 16 print(f"B (numerical) = {B_numerical}")

Let me try to get exact value

R_f(tau) for f(t) = cos(t) on [-pi/2, pi/2]

[/PYTHON]

[THOUGHT] Let me compute the exact values symbolically. I need to find the exact value of B.

For the autocorrelation $R_f(\tau)$, I need to compute it analytically. Let me work this out:

For $f(t) = \cos(t)$ on $[-\pi/2, \pi/2]$:

$R_f(\tau) = \int f(t) f(t-\tau) dt$

For $0 \leq \tau \leq \pi$, the overlap region is $[\tau - \pi/2, \pi/2]$:

$R_f(\tau) = \int_{\tau-\pi/2}^{\pi/2} \cos(t)\cos(t-\tau) dt$

Using $\cos(t)\cos(t-\tau) = \frac{1}{2}[\cos(\tau) + \cos(2t-\tau)]$:

$R_f(\tau) = \frac{1}{2}\int_{\tau-\pi/2}^{\pi/2} [\cos(\tau) + \cos(2t-\tau)] dt$

$= \frac{1}{2}[\cos(\tau)(\pi - \tau) + \frac{1}{2}\sin(2t-\tau)|_{\tau-\pi/2}^{\pi/2}]$

$= \frac{1}{2}[\cos(\tau)(\pi - \tau) + \frac{1}{2}(\sin(\pi-\tau) - \sin(\tau - \pi))]$

$= \frac{1}{2}[\cos(\tau)(\pi - \tau) + \sin(\tau)]$

So $R_f(\tau) = \frac{(\pi-\tau)\cos(\tau) + \sin(\tau)}{2}$ for $0 \leq \tau \leq \pi$.

By symmetry, $R_f(\tau) = R_f(-\tau)$.

Now I need $\int_{-\pi}^{\pi} R_f(\tau)^2 d\tau = 2\int_0^{\pi} R_f(\tau)^2 d\tau$

$= 2\int_0^{\pi} \frac{[(\pi-\tau)\cos(\tau) + \sin(\tau)]^2}{4} d\tau$

$= \frac{1}{2}\int_0^{\pi} [(\pi-\tau)\cos(\tau) + \sin(\tau)]^2 d\tau$

The numerical value was approximately 4.547, and $\frac{3\pi^2}{16} + \frac{\pi}{4} \approx 1.85 + 0.785 \approx 2.64$, which doesn't match.

Let me trust the numerical computation:

• $A = \pi^2/4$
• $B \approx 1.7857...$

Looking at $B = \frac{3\pi^2 + 8}{16\pi} \cdot \pi = \frac{3\pi^2 + 8}{16}$? Let me check: $(3\pi^2 + 8)/16 \approx (29.6 + 8)/16 \approx 2.35$. Not quite.

Actually, from the numerical: $B \approx 1.7857 \approx \frac{5}{28} \cdot \pi^2$? Let's see: $\pi^2/16 \cdot 1.8 \approx 1.11$.

Let me check if $B = \frac{5\pi^2}{28}$: $5 \times 9.87/28 \approx 1.76$. Close!

Or $B = \frac{9\pi^2}{16 \times 3.1} \approx 1.78$.

Actually $B = \frac{9\pi^2}{16 \times \pi} = \frac{9\pi}{16} \approx 1.767$. Very close!

Let me compute $\mathcal{K}$:

• $F(0) = 2$, so $(F(0))^4 = 16$
• $A = \pi^2/4$
• $B \approx 1.7857$ (let me use the numerical value)

$\mathcal{K} = 37 \times 16 + 999(96B - 16A^2 - 30A) + 123456$

$= 592 + 999(96 \times 1.7857 - 16 \times (\pi^2/4)^2 - 30 \times \pi^2/4) + 123456$

$= 592 + 999(171.43 - 16 \times \pi^4/16 - 7.5\pi^2) + 123456$

$= 592 + 999(171.43 - \pi^4 - 7.5\pi^2) + 123456$

$= 592 + 999(171.43 - 97.41 - 74.02) + 123456$

$= 592 + 999 \times 0 + 123456 = 124048$

So $\mathcal{K} = 124048$. [/THOUGHT]

\boxed{124048}