[THOUGHT] I'll compute the residue at the pole z = 3i using Sympy. The integral over the real line will be 2πi times the residue in the upper half-plane, since e^{iz} decays there. The residue for a pole of order 8 at z=3i of f(z) = P(z)e^{iz}/(z-3i)^8 is (1/7!) times the 7th derivative of P(z)e^{iz} evaluated at z=3i. Then I = 2πi * residue. Finally, extract real and imaginary parts (a, b), compute floor(10a) + floor(b). [/THOUGHT] [PYTHON] import sympy as sp

Define variable and polynomial

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

Define g(z) = P(z) * e^{iz}

g = P * sp.exp(sp.I*z)

7th derivative at z = 3i

g7 = sp.diff(g, z, 7) g7_at_3i = sp.simplify(g7.subs(z, 3*sp.I))

Residue = g^{(7)}(3i) / 7!

residue = sp.simplify(g7_at_3i / sp.factorial(7))

Integral I = 2pii * residue

I = sp.simplify(2sp.pisp.I*residue)

Get real and imaginary parts

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

a_simpl = sp.simplify(a) b_simpl = sp.simplify(b)

print("I =", I) print("a =", a_simpl) print("b =", b_simpl)

Compute floors

from sympy import floor val = floor(10*a_simpl) + floor(b_simpl) print("Result =", sp.simplify(val)) [/PYTHON]

[THOUGHT] Using the residue calculation and the Python output, the integral evaluates to (I = a + bi) with (a = \frac{134\pi e^{-3}}{35}) and (b = \frac{5479\pi e^{-3}}{420}). Python then computed (\lfloor 10a \rfloor + \lfloor b \rfloor = 7). No further computation is needed. [/THOUGHT] \boxed{7}