[THOUGHT] We need to evaluate the integral [ I = \int_{-\infty}^{\infty} \frac{P(x) e^{ix}}{(x - 3i)^{8}} dx ] where (P(x)) is a polynomial of degree 7. The integrand is analytic except at (x = 3i) (pole of order 8). Since the integration path is along the real axis, we can close the contour in the upper half-plane if the integrand decays sufficiently for large (|x|) in the upper half-plane. The factor (e^{ix}) decays in the upper half-plane because (e^{ix} = e^{i(x+iy)} = e^{ix - y}) with (y>0). So we can close the contour upward. Then by the residue theorem, (I = 2\pi i \cdot \text{Residue at } x = 3i) (since the pole is inside the contour? Wait: (x = 3i) is in the upper half-plane (positive imaginary part), yes, because (3i) has imaginary part 3 > 0. So indeed the pole is inside the closed contour (upper half-plane). So [ I = 2\pi i \cdot \text{Res}{x=3i} \frac{P(x) e^{ix}}{(x - 3i)^{8}}. ] Since the denominator is ((x-3i)^8), the residue for a pole of order (n) is given by [ \text{Res}{z=z_0} f(z) = \frac{1}{(n-1)!} \lim_{z\to z_0} \frac{d^{n-1}}{dz^{n-1}} \left[ (z-z_0)^n f(z) \right]. ] Here (n=8), so [ \text{Res}{x=3i} = \frac{1}{7!} \lim{x\to 3i} \frac{d^7}{dx^7} \left[ (x-3i)^8 \cdot \frac{P(x) e^{ix}}{(x-3i)^{8}} \right] = \frac{1}{7!} \lim_{x\to 3i} \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right]. ] Thus [ I = 2\pi i \cdot \frac{1}{7!} \left. \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right] \right|_{x=3i}. ] We need to compute the 7th derivative of (P(x) e^{ix}) evaluated at (x=3i). Since (P(x)) is a polynomial of degree 7, its 7th derivative will be constant (the leading term's derivative). Let's compute.

Let (f(x) = P(x) e^{ix}). Use Leibniz rule: [ \frac{d^7}{dx^7} [P(x) e^{ix}] = \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(x) \cdot (e^{ix})^{(7-k)}. ] But ((e^{ix})^{(m)} = i^m e^{ix}). So [ = \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(x) i^{7-k} e^{ix}. ] Evaluate at (x=3i).

Alternatively, note that (P(x)) is degree 7, so (P^{(7)}(x) = 7! \cdot) leading coefficient. The leading term of (P(x)) is (x^7) (coefficient 1). So (P^{(7)}(x) = 7!). Lower derivatives depend on polynomial coefficients.

We can compute explicitly using Python symbolic differentiation.

We'll compute (P(x)) as given, then compute the 7th derivative of (P(x) e^{ix}) symbolically and evaluate at (x=3i). Then multiply by (2\pi i / 7!) to get (I). Then extract real and imaginary parts (a) and (b), compute floor(10a) and floor(b), and sum them.

Let's do it step by step in Python. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

x = sp.symbols('x') i = sp.I # imaginary unit

Define polynomial P(x)

P = (x7 - 14ix6 + (85 - 60*i)x**5 - (240 + 210i)x**4 + (420 - 840i)x**3 - (560 + 1680i)x**2 + (1344 - 2688i)*x + 3840)

Define f(x) = P(x) * exp(i*x)

f = P * sp.exp(i*x)

Compute 7th derivative

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

Evaluate at x = 3i

val = sp.simplify(f7.subs(x, 3*i)) print("7th derivative at 3i:", val)

Compute residue: val / 7!

residue = val / sp.factorial(7) print("Residue:", residue)

Compute integral I = 2pii * residue

I = 2sp.pii * residue print("Integral I:", I) print("I simplified:", sp.simplify(I))

Extract real and imaginary parts

a = sp.re(I) b = sp.im(I) print("a =", a) print("b =", b)

Compute floor(10a) and floor(b)

floor_10a = sp.floor(10*a) floor_b = sp.floor(b) print("floor(10a) =", floor_10a) print("floor(b) =", floor_b)

Compute sum

result = floor_10a + floor_b print("Result:", result) [/PYTHON]

[THOUGHT] From the Python output, we have:

• ( a = \frac{134\pi e^{-3}}{35} )
• ( b = \frac{5479\pi e^{-3}}{420} )
• ( \lfloor 10a \rfloor = 5 )
• ( \lfloor b \rfloor = 2 )
• Result = ( 5 + 2 = 7 )

Thus the required integer is 7. [/THOUGHT] \boxed{7}