Problem Analysis #1

[THOUGHT] The integral is evaluated using the residue theorem. The pole is at ( z_0 = \frac{205}{69} i ) in the upper half-plane. Closing the contour there gives ( I = 2\pi i ) times the residue of ( \frac{P(z) e^{i z}}{(z - z_0)^8} ) at ( z_0 ), which is ( \frac{1}{7!} \frac{d^7}{dz^7} [P(z) e^{i z}] \big|_{z=z_0} ).

Use SymPy to define P(z), compute Q(z) = P(z) * exp(Iz), differentiate 7 times, evaluate at z_0, divide by 7!, multiply by 2pi*I, then extract real and imaginary parts numerically to compute floor(10a) + floor(b). [/THOUGHT] [PYTHON] from sympy import *

z = symbols('z') I_sym = I pole = Rational(205,69) * I_sym

P = (z7 -14*I_sym * z6 + (85 -60*I_sym)z**5 + (-240 -210I_sym)z**4 + (420 -840I_sym)z**3 + (-560 -1680I_sym)z**2 + (1344 -2688I_sym)*z + 3840)

Q = P * exp(I_sym * z) d7Q = diff(Q, z, 7) res = d7Q.subs(z, pole) / factorial(7) I_int = 2 * pi * I_sym * res

a = re(I_int).evalf(30) b = im(I_int).evalf(30)

print('a:', a) print('b:', b) print('floor(10a):', floor(10a)) print('floor(b):', floor(b)) print('sum:', floor(10a) + floor(b)) [/PYTHON]

Explanation

The reasoning follows the residue theorem for contour integration. The integral ( I = \int_{-\infty}^{\infty} \frac{P(x)

[THOUGHT] The Python output gives floor(10a) = 7 and floor(b) = 2, so their sum is 9. [/THOUGHT] \boxed{9}

Explanation

The integral evaluation relies on the residue theorem, as the integrand has a pole of order 8 at ( z_0 = \frac{205}{69} i ) in the upper half-plane. For ( \int_{-\infty}^{\infty} f(x) , dx ) with ( e^{i x} ), closing the contour in the upper half-plane yields ( I = 2\pi i \times \text{Res}(f, z_0) ).

The residue for a pole of order 8 is ( \text{Res} = \frac{1}{7!} \frac{d^7}{dz^7} \left[ P(z) e^{i z} \right] \bigg|_{z=z_0} ).

SymPy computes this exactly:

• Defines ( P(z) ) as the given polynomial.
• Forms ( Q(z) = P(z) e^{i z} ).
• Differentiates ( Q(z) ) seven times.
• Evaluates at ( z_0 ), divides by ( 7! ), and multiplies by ( 2\pi i ).

Numerical evaluation gives ( I \approx 0.7208 + 2.2479i ), so ( a \approx 0.7208 ), ( b \approx 2.2479 ). Thus, ( \lfloor 10a \rfloor = 7 ), ( \lfloor b \rfloor = 2 ), and their sum is 9.