[THOUGHT] I will use sympy to:

1. Compute F(0)=∫_{-π/2}^{π/2} cos(t) dt for a=1.
2. Evaluate the two definite integrals A and B exactly.
3. Form the expression for K and simplify it to an exact integer.

I will print all intermediate symbolic results to verify them, then read off the final integer value of K. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

x = sp.symbols('x', real=True) t = sp.symbols('t', real=True)

1) Compute F(0) for a=1: f(t) = cos(t) on [-pi/2, pi/2]

F0 = sp.integrate(sp.cos(t), (t, -sp.pi/2, sp.pi/2)) print("F(0) =", sp.simplify(F0))

2) Compute A and B

A = sp.integrate(sp.cos(sp.pix/2)2 / (x2 - 1)**2, (x, -sp.oo, sp.oo)) B = sp.integrate(sp.cos(sp.pix/2)4 / (x2 - 1)**4, (x, -sp.oo, sp.oo))

print("A =", sp.simplify(A)) print("B =", sp.simplify(B))

3) Compute K

K = 37F0**4 + 999(96B - 16A**2 - 30*A) + 123456 K_simplified = sp.simplify(K) print("K simplified =", K_simplified) print("K as integer =", int(K_simplified)) [/PYTHON]

[THOUGHT] I need to correct for the integration error by switching from symbolic integration (which raised the MellinTransformStripError) to high-precision numerical integration using mpmath, then compute K numerically and identify the exact integer. [/THOUGHT] [PYTHON] import sympy as sp

High precision numerical work with mpmath

mp = sp.mpmath mp.mp.dps = 80 # set high precision

Symbol for t to compute F(0)

t = sp.symbols('t', real=True)

1) Compute F(0) for a=1: f(t) = cos(t) on [-pi/2, pi/2]

F0 = sp.integrate(sp.cos(t), (t, -sp.pi/2, sp.pi/2)) print("F(0) symbolic =", F0) F0_num = mp.mpf(sp.N(F0, 50)) print("F(0) numeric =", F0_num)

2) Define integrands for A and B, handling removable singularities at x = ±1

def fA(x): x = mp.mpf(x) if mp.fabs(x-1) < mp.mpf('1e-8') or mp.fabs(x+1) < mp.mpf('1e-8'): return (mp.pi**2) / 16 return (mp.cos(mp.pi*x/2)2) / (x2 - 1)**2

def fB(x): x = mp.mpf(x) if mp.fabs(x-1) < mp.mpf('1e-8') or mp.fabs(x+1) < mp.mpf('1e-8'): return (mp.pi**4) / 256 return (mp.cos(mp.pi*x/2)4) / (x2 - 1)**4

3) Compute A and B using numerical integration, exploiting evenness

A = ∫{-∞}^{∞} ... dx = 2 ∫{0}^{∞} ... dx

A_num = 2 * mp.quad(fA, [0, 1-mp.mpf('1e-6'), 1+mp.mpf('1e-6'), mp.inf]) B_num = 2 * mp.quad(fB, [0, 1-mp.mpf('1e-6'), 1+mp.mpf('1e-6'), mp.inf])

print("A numeric =", A_num) print("B numeric =", B_num)

4) Compute K numerically

K_num = 37*(F0_num4) + 999*(96B_num - 16(A_num2) - 30*A_num) + 123456 print("K numeric (high precision) =", K_num)

5) Round K to nearest integer and also use nsimplify for verification

K_int = int(mp.nint(K_num)) # nearest integer print("K as nearest integer =", K_int)

K_nsimplified = sp.nsimplify(str(K_num), [sp.pi]) print("nsimplified K =", K_nsimplified)

[THOUGHT] Using the Fourier-transform relationships and Parseval/Wiener–Khinchin:

• For (a=1), (F(\omega)=\displaystyle\int_{-\pi/2}^{\pi/2}\cos t,e^{-i\omega t}dt = \frac{2\cos\bigl(\frac{\pi\omega}{2}\bigr)}{1-\omega^2}), so (F(0)=2).

• Then (|F(\omega)|^2 = \dfrac{4\cos^2(\pi\omega/2)}{(\omega^2-1)^2}), so [ A=\int_{-\infty}^{\infty}\frac{\cos^2\bigl(\frac{\pi x}{2}\bigr)}{(x^2-1)^2}dx =\frac14\int_{-\infty}^{\infty}|F(\omega)|^2d\omega =\frac{\pi^2}{4}. ]

• Since (|F(\omega)|^4 = \dfrac{16\cos^4(\pi\omega/2)}{(\omega^2-1)^4}), [ B = \int_{-\infty}^{\infty} \frac{\cos^4\bigl(\frac{\pi x}{2}\bigr)}{(x^2-1)^4}dx =\frac1{16}\int_{-\infty}^{\infty}|F(\omega)|^4d\omega =\frac{\pi}{8}\int_{-\infty}^{\infty}R_f(\tau)^2d\tau. ] For (a=1), (R_f(\tau)=\frac12\bigl[(\pi-|\tau|)\cos\tau+\sin|\tau|\bigr]) for (|\tau|\le\pi), and 0 otherwise. Computing (\int R_f^2) gives [ B = \frac{\pi^4}{96} + \frac{5\pi^2}{64}. ]

Then [ A=\frac{\pi^2}{4},\quad A^2=\frac{\pi^4}{16},\quad 96B = \pi^4 + \frac{15}{2}\pi^2, ] so [ 96B - 16A^2 = \frac{15}{2}\pi^2,\quad -30A = -\frac{15}{2}\pi^2,\quad \Rightarrow 96B - 16A^2 - 30A = 0. ]

Thus [ \mathcal{K} = 37(F(0))^4 + 999\cdot 0 + 123456 = 37\cdot 2^4 + 123456 = 592 + 123456 = 124048. ] [/THOUGHT] \boxed{124048}