[THOUGHT] We need to evaluate the integral: [ I = \int_{-\infty}^{\infty} \frac{P(x) e^{ix}}{\left(x - \frac{205}{69} i\right)^{8}} , dx ] where (P(x)) is a degree-7 polynomial with complex coefficients. The integrand has a pole at (x = \frac{205}{69} i) in the complex plane. Since the exponential is (e^{ix} = e^{i \cdot (Re(x) + i Im(x))} = e^{i Re(x)} e^{-Im(x)}). For convergence of the integral along the real axis, we need the factor (e^{ix}) to decay as (|x| \to \infty) in the upper or lower half-plane. Actually, (e^{ix} = e^{i x}) with (x) real: (e^{ix} = \cos x + i \sin x), magnitude 1. So it doesn't decay. Wait, but the integral is over real (x), so the integrand is (P(x) e^{ix} / (x - i c)^8) where (c = 205/69). The denominator grows as (|x|^8) for large real (x), so the integral converges absolutely because polynomial numerator times (e^{ix}) (bounded) divided by (|x|^8) yields decay like (1/|x|^8) (since degree of numerator is 7, denominator degree 8). So it's absolutely convergent.

But to evaluate, we can consider contour integration in the complex plane. The integrand is analytic except at the pole (z = i c) where (c = 205/69). The exponential factor (e^{iz}) decays in the upper half-plane if (Im(z) > 0) because (e^{iz} = e^{i (Re z + i Im z)} = e^{i Re z} e^{-Im z}). So for (Im(z) > 0), magnitude (e^{-Im z}) decays. Since our pole is at (Im(z) = c > 0), it lies in the upper half-plane. To apply the residue theorem, we close the contour in the upper half-plane (where (e^{iz}) decays). However, we must check if the semicircular arc at infinity contributes zero. For large (|z|) in the upper half-plane, (|e^{iz}| = e^{-Im z} \le 1) if (Im z \ge 0). But for the arc with (z = R e^{i\theta}), (0 \le \theta \le \pi), we have (Im z = R \sin\theta \ge 0), so (|e^{iz}| = e^{-R \sin\theta} \le 1). However, the denominator grows as (|z|^8), so the integrand decays as (1/R^8) times bounded exponential. So the arc contribution vanishes as (R \to \infty). Therefore, [ \int_{-\infty}^{\infty} \frac{P(z) e^{iz}}{(z - i c)^8} dz = 2\pi i \cdot \operatorname{Res}{z = i c} \frac{P(z) e^{iz}}{(z - i c)^8}. ] The residue at a pole of order 8 is given by: [ \operatorname{Res}{z = a} \frac{f(z)}{(z-a)^n} = \frac{1}{(n-1)!} f^{(n-1)}(a), ] where (f(z) = P(z) e^{iz}). Here (n=8), (a = i c). So: [ \operatorname{Res}{z = i c} \frac{P(z) e^{iz}}{(z - i c)^8} = \frac{1}{7!} \frac{d^{7}}{dz^{7}} \left[ P(z) e^{iz} \right]{z = i c}. ] Thus, [ I = 2\pi i \cdot \frac{1}{7!} \left. \frac{d^{7}}{dz^{7}} \left( P(z) e^{iz} \right) \right|_{z = i c}. ] We need to compute (P(z) e^{iz}), take its 7th derivative, evaluate at (z = i c), multiply by (2\pi i / 7!), then find (a = \Re(I)), (b = \Im(I)), compute (\lfloor 10a \rfloor + \lfloor b \rfloor).

We can do this with symbolic computation in Python using sympy. However, derivatives of order 7 might be heavy but doable. Let's define (c = 205/69). Compute (P(z)) as given, then compute derivative.

But note: (P(z)) is a polynomial with complex coefficients. We'll compute exactly using sympy rationals? Might be messy. Let's use sympy's derivative and simplify.

We'll compute (f(z) = P(z) * exp(iz)). Then compute (f7 = diff(f, z, 7)). Evaluate at (z = ic). Then compute (I = 2pii/factorial(7) * f7|_{z=i*c}).

We need numeric values for a and b. Use high precision? Might be large numbers. Use mpmath or sympy's N with high precision.

Let's implement.

First, import sympy, define symbols, define P(x) as given, compute derivative, substitute, evaluate numerically.

We need to be careful with complex i. Use sympy.I.

Let's code. </think>

[PYTHON] import sympy as sp

Define symbols

x = sp.symbols('x', complex=True) I_sym = sp.I

Define c = 205/69

c = sp.Rational(205, 69)

Define P(x) as given

P = (x7 - 14I_symx6 + (85 - 60*I_sym)x**5 - (240 + 210I_sym)x**4 + (420 - 840I_sym)x**3 - (560 + 1680I_sym)x**2 + (1344 - 2688I_sym)*x + 3840)

Define f(z) = P(z) * exp(I*z)

z = sp.symbols('z', complex=True) f = P.subs(x, z) * sp.exp(I_sym * z)

Compute 7th derivative

f7 = sp.diff(f, z, 7)

Evaluate at z = I*c

point = I_sym * c f7_at_point = f7.subs(z, point)

Compute I = 2Ipi / 7! * f7_at_point

I_val = (2 * sp.pi * I_sym / sp.factorial(7)) * f7_at_point

Simplify (might be heavy)

I_simplified = sp.simplify(I_val)

Convert to numeric (high precision)

I_numeric = sp.N(I_simplified, 30) # 30 digits

Extract real and imaginary parts

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

print("I =", I_numeric) print("Real part a =", a) print("Imaginary part b =", b) print("Floor(10a) =", sp.floor(10a)) print("Floor(b) =", sp.floor(b)) print("Floor(10a) + Floor(b) =", sp.floor(10a) + sp.floor(b)) [/PYTHON]

[THOUGHT] The Python output gives ( a \approx 0.7208218847574711 ) and ( b \approx 2.2479236808971603 ).
Then ( \lfloor 10a \rfloor = \lfloor 7.208218847574711 \rfloor = 7 ) and ( \lfloor b \rfloor = \lfloor 2.2479236808971603 \rfloor = 2 ).
Sum = ( 7 + 2 = 9 ).

\boxed{9}