[THOUGHT] Use the residue theorem with a contour in the upper half-plane. The integrand has an 8th-order pole at (z_0 = \frac{205}{69}i) in the upper half-plane and decays on the semicircle due to (e^{iz}). Thus, [ I = \int_{-\infty}^{\infty} \frac{P(x)e^{ix}}{(x-\frac{205}{69}i)^8}dx = 2\pi i \cdot \text{Res}\left(\frac{P(z)e^{iz}}{(z-\frac{205}{69}i)^8}, z=z_0\right). ] For an order-8 pole, [ \text{Res} = \frac{1}{7!}\frac{d^7}{dz^7}\left[P(z)e^{iz}\right]_{z=z_0}. ] Compute this with sympy, then get real and imaginary parts of (I), then (\lfloor 10a \rfloor + \lfloor b \rfloor). [/THOUGHT] [PYTHON] import sympy as sp

Define symbol and imaginary unit

z = sp.symbols('z') I = sp.I

Define polynomial P(z)

P = (z7 - 14Iz6 + (85-60*I)z**5 - (240+210I)z**4 + (420-840I)z**3 - (560+1680I)z**2 + (1344-2688I)*z + 3840)

Define z0

z0 = 205*I/69

Define the function whose 7th derivative we need

f = Psp.exp(Iz)

Compute 7th derivative

d7 = sp.diff(f, z, 7)

Residue at z0 for 8th-order pole

res = d7.subs(z, z0) / sp.factorial(7)

Integral I via residue theorem

integral = 2sp.piI*res

Simplify

integral_simplified = sp.simplify(integral)

Get real and imaginary parts with high precision

a = sp.re(integral_simplified) b = sp.im(integral_simplified)

a_n = sp.N(a, 50) b_n = sp.N(b, 50)

print("Integral I =", integral_simplified) print("a (real part) ~", a_n) print("b (imag part) ~", b_n)

Compute the requested quantity

val = int(sp.floor(10*a_n) + sp.floor(b_n)) print("Result =", val)

[THOUGHT] Using the residue computation from the Python output, the integral’s real and imaginary parts are (a \approx 0.7208) and (b \approx 2.2479). Thus (\lfloor 10a \rfloor = \lfloor 7.208...\rfloor = 7) and (\lfloor b \rfloor = \lfloor 2.247...\rfloor = 2), giving (7+2=9). [ \boxed{9} ]