[THOUGHT] The given integral involves a polynomial ( P(x) ) and a complex exponential ( e^{ix} ) divided by a power of a complex linear term ( (x - 3i)^8 ). To evaluate this integral, I can use the residue theorem from complex analysis. Specifically, the integral along the real axis can be computed by considering a closed contour in the upper half-plane, which includes the pole at ( x = 3i ).

Here, the integrand ( \frac{P(x) e^{ix}}{(x-3i)^8} ) has a pole of order 8 at ( x = 3i ). According to the residue theorem, we need to find the residue of the integrand at ( x = 3i ). The residue for a pole of order ( n ) is given by:

[ \text{Res}\left(\frac{P(x) e^{ix}}{(x-3i)^8}, 3i\right) = \frac{1}{(8-1)!} \left. \frac{d^{7}}{dx^{7}} \left[ (x-3i)^8 \cdot \frac{P(x) e^{ix}}{(x-3i)^8} \right] \right|{x=3i} = \frac{1}{7!} \left. \frac{d^{7}}{dx^{7}} \left[ P(x) e^{ix} \right] \right|{x=3i}. ]

Let's compute the 7th derivative of ( P(x) e^{ix} ).

We have: [ P(x) = x^7 - 14ix^6 + (85-60i)x^5 - (240+210i)x^4 + (420-840i)x^3 - (560+1680i)x^2 + (1344-2688i)x + 3840. ]

The 7th derivative of ( P(x) e^{ix} ) is: [ \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right] = \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(x) \left( e^{ix} \right)^{(7-k)}. ]

Calculating the derivatives of ( P(x) ): [ P'(x) = 7x^6 - 84ix^5 + 5(85-60i)x^4 - 4(240+210i)x^3 + 3(420-840i)x^2 - 2(560+1680i)x + (1344-2688i), ] [ P''(x) = 42x^5 - 420ix^4 + 20(85-60i)x^3 - 12(240+210i)x^2 + 6(420-840i)x - 2(560+1680i), ] [ P'''(x) = 210x^4 - 1680ix^3 + 60(85-60i)x^2 - 24(240+210i)x + 6(420-840i), ] [ P^{(4)}(x) = 840x^3 - 5040ix^2 + 120(85-60i)x - 24(240+210i), ] [ P^{(5)}(x) = 2520x^2 - 10080ix + 120(85-60i), ] [ P^{(6)}(x) = 5040x - 10080i, ] [ P^{(7)}(x) = 5040. ]

Now, calculating the derivatives of ( e^{ix} ): [ \left( e^{ix} \right)' = ie^{ix}, \quad \left( e^{ix} \right)'' = -e^{ix}, \quad \left( e^{ix} \right)''' = -ie^{ix}, \quad \left( e^{ix} \right)^{(4)} = e^{ix}, \quad \left( e^{ix} \right)^{(5)} = ie^{ix}, \quad \left( e^{ix} \right)^{(6)} = -e^{ix}, \quad \left( e^{ix} \right)^{(7)} = -ie^{ix}. ]

Substituting ( x = 3i ) and evaluating: [ \left. \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right] \right|_{x=3i} = 5040(-ie^{i(3i)}) + 7 \cdot 5040(-e^{i(3i)}) + 21 \cdot 10080i(e^{i(3i)}) + 35 \cdot 120(85-60i)(-ie^{i(3i)}) + 35 \cdot 120(240+210i)(-e^{i(3i)}) + 21 \cdot 5040(-ie^{i(3i)}) + 7 \cdot 840(e^{i(3i)}) + 42(-ie^{i(3i)}). ]

First, calculate ( e^{i(3i)} = e^{-3} ): [ e^{i(3i)} = e^{-3}. ]

Substitute ( e^{-3} ) into the expression: [ \left. \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right] \right|_{x=3i} = 5040(-i e^{-3}) + 7 \cdot 5040(-e^{-3}) + 21 \cdot 10080i(e^{-3}) + 35 \cdot 120(85-60i)(-ie^{-3}) + 35 \cdot 120(240+210i)(-e^{-3}) + 21 \cdot 5040(-ie^{-3}) + 7 \cdot 840(e^{-3}) + 42(-ie^{-3}). ]

Simplify: [ = -5040ie^{-3} - 35280e^{-3} + 210576ie^{-3} - 35 \cdot 120 \cdot 85ie^{-3} + 35 \cdot 120 \cdot 60 e^{-3} - 35 \cdot 120 \cdot 240 e^{-3} - 35 \cdot 120 \cdot 210 i e^{-3} - 105680ie^{-3} + 5880e^{-3} - 42ie^{-3}. ]

Combine like terms: [ = (-5040i + 210576i - 35 \cdot 120 \cdot 85i - 35 \cdot 120 \cdot 210i - 105680i - 42i)e^{-3} + (-35280 + 35 \cdot 120 \cdot 60 - 35 \cdot 120 \cdot 240 + 5880)e^{-3}. ]

Calculate the coefficients: [ -5040i + 210576i - 35 \cdot 120 \cdot 85i - 35 \cdot 120 \cdot 210i - 105680i - 42i = -5040i + 210576i - 35700i - 85800i - 105680i - 42i = -42 \cdot 120 \cdot 210i = -10582i, ] [ -35280 + 35 \cdot 120 \cdot 60 - 35 \cdot 120 \cdot 240 + 5880 = -35280 + 25200 - 100800 + 5880 = -10200. ]

Thus, the residue is: [ \text{Res} = \frac{1}{7!} (-10582ie^{-3} - 10200e^{-3}) = \frac{-10582i - 10200}{5040e^3}. ]

The integral is ( 2\pi i ) times the residue: [ I = 2\pi i \cdot \frac{-10582i - 10200}{5040e^3} = 2\pi \cdot \frac{-10582i^2 - 10200i}{5040e^3} = 2\pi \cdot \frac{10582 - 10200i}{5040e^3} = \frac{2\pi (10582 - 10200i)}{5040e^3}. ]

Simplify: [ I = \frac{\pi (10582 - 10200i)}{2520e^3} = \frac{10582\pi}{2520e^3} - \frac{10200\pi i}{2520e^3} = \frac{10582\pi}{2520e^3} - \frac{5\pi i}{e^3}. ]

Calculate the real and imaginary parts: Using Python, compute the numerical values of ( \frac{10582\pi}{2520e^3} ) and ( -\frac{5\pi}{e^3} ). [/THOUGHT] [PYTHON] import math

Constants

pi = math.pi e = math.e

Real part

real_part = (10582 * pi) / (2520 * e**3)

Imaginary part

imaginary_part = -(5 * pi) / e**3

print(f"Real part: {real_part}") print(f"Imaginary part: {imaginary_part}")

[THOUGHT] Given the numerical values from the Python output:

• Real part ( a = 0.6568007550918197 )
• Imaginary part ( b = -0.7820534411412708 )

We need to compute ( \lfloor 10a \rfloor + \lfloor b \rfloor ).

First, calculate ( 10a ): [ 10a \approx 10 \times 0.6568007550918197 = 6.568007550918197 ]

Then, take the floor of ( 10a ): [ \lfloor 10a \rfloor = \lfloor 6.568007550918197 \rfloor = 6 ]

Next, take the floor of ( b ): [ \lfloor b \rfloor = \lfloor -0.7820534411412708 \rfloor = -1 ]

Finally, compute the sum: [ \lfloor 10a \rfloor + \lfloor b \rfloor = 6 + (-1) = 5 ]

The final answer is: [ \boxed{5} ]