Problem Analysis #2

[THOUGHT] The integral is evaluated using contour integration in the upper half-plane, enclosing the pole of order 8 at z = 3i. The value is I = 2πi × Res[ P(z) e^{i z} / (z - 3i)^8 , z=3i ]. The residue for a pole of order 8 is (1/7!) × d^7/dz^7 [P(z) e^{i z}] at z=3i. Use SymPy to compute this exactly, then evaluate numerically for a and b. [/THOUGHT] [PYTHON] from sympy import *

z = symbols('z') P = (z7 - 14Iz6 + (85 - 60*I)z**5 - (240 + 210I)z**4 + (420 - 840I)z**3 - (560 + 1680I)z**2 + (1344 - 2688I)*z + 3840) Q = P * exp(I * z) z0 = 3 * I Q7 = diff(Q, z, 7).subs(z, z0) res = simplify(Q7 / factorial(7)) I_val = simplify(2 * pi * I * res) print(I_val) print(I_val.evalf(30)) a = re(I_val).evalf(30) b = im(I_val).evalf(30) print(float(a), float(b)) [/PYTHON]

[THOUGHT] The integral evaluates to approximately 0.5988294920738872 + 2.0404146685776294 i. Thus, a ≈ 0.5988294920738872, b ≈ 2.0404146685776294. floor(10a) = floor(5.988294920738872) = 5, floor(b) = 2. Therefore, 5 + 2 = 7. [/THOUGHT] \boxed{7}

Explanation

The reasoning follows the residue theorem for contour integration. The integral I is evaluated over the real line using a semicircular contour in the upper half-plane, which encloses the pole of order 8 at z = 3i. Thus, I = 2πi × Res[ P(z) e^{i z} / (z - 3i)^8 , z=3i ].

The residue for a pole of order 8 is given by (1/(8-1)!) × [d^7/dz^7 (P(z) e^{i z})] evaluated at z = 3i, or (1/7!) × seventh derivative of Q(z) = P(z) e^{i z} at z = 3i.

SymPy computes this exactly: define P(z) as the given polynomial, form Q(z) = P(z) exp(I z), compute the seventh derivative of Q(z), evaluate at z0 = 3I, divide by 7!, and multiply by 2πi, yielding pi*(-1870 + 1739I)(1 - 2*I)*exp(-3)/420.

Numerical evaluation provides I ≈ 0.5988294920738872 + 2.0404146685776294 i, so a ≈ 0.5988294920738872 and b ≈ 2.0404146685776294.

Then, compute 10a ≈ 5.988294920738872, so floor(10a) = 5. For b ≈ 2.0404146685776294, floor(b) = 2. The sum is 5 + 2 = 7.