The problem asks us to evaluate an integral $I$ involving a polynomial $Q(x)$ determined by the roots $a, b, c$, which satisfy a system of power sums.

First, we determine the coefficients of the polynomial $Q(x) = (x-a)(x-b)(x-c) = x^3 - e_1 x^2 + e_2 x - e_3$. Using Newton sums: $e_1 = p_1 = 151$ $e_2 = \frac{e_1 p_1 - p_2}{2}$ $e_3 = \frac{e_2 p_1 - e_1 p_2 + p_3}{3}$

Then we evaluate the integral $I = \int_{-\infty}^{\infty} \frac{Q(x)e^{ix}}{(x-z_0)^4} dx$ using the residue theorem, where $z_0 = \frac{205}{69}i$. Since $z_0$ is in the upper half-plane, the integral is $2\pi i \cdot \text{Res}(f, z_0)$. The residue for a pole of order 4 is $\frac{1}{6} \frac{d^3}{dz^3} [Q(z)e^{iz}] \bigg|_{z=z_0}$.

Let's perform the calculations.

[PYTHON] import cmath

Given power sums

p1 = 151 p2 = 10939 p3 = 957871

Calculate elementary symmetric polynomials

e1 = p1 e2 = (e1 * p1 - p2) // 2 e3 = (e2 * p1 - e1 * p2 + p3) // 3

print(f"{e1=}, {e2=}, {e3=}")

Define Q and its derivatives at a point z

Q(z) = z^3 - e1z^2 + e2z - e3

def get_Q_derivs(z): # Q(z) Q0 = z3 - e1*z2 + e2z - e3 # Q'(z) = 3z^2 - 2e1z + e2 Q1 = 3z**2 - 2e1z + e2 # Q''(z) = 6z - 2e1 Q2 = 6z - 2*e1 # Q'''(z) = 6 Q3 = 6 return Q0, Q1, Q2, Q3

The pole z0

z0 = 205/69 * 1j

Calculate the term from the 3rd derivative of Q(z)e^{iz}

g'''(z) = [Q'''(z) + 3i Q''(z) - 3 Q'(z) - i Q(z)] e^{iz}

Q0, Q1, Q2, Q3 = get_Q_derivs(z0) term_bracket = Q3 + 3j * Q2 - 3 * Q1 - 1j * Q0 exp_term = cmath.exp(1j * z0)

Integral I = 2pii * (1/6) * term_bracket * exp_term

I = (2 * cmath.pi * 1j / 6) * term_bracket * exp_term

alpha = I.real beta = I.imag

print(f"{I=}") print(f"{alpha=}") print(f"{beta=}")

Compute final answer

ans = int(10 * alpha) + int(beta) # Using int() acts like floor for positive, but check signs import math ans_floor = math.floor(10 * alpha) + math.floor(beta) print(f"{ans_floor=}") [/PYTHON]

[THOUGHT] The Python code calculated the elementary symmetric polynomials $e_1, e_2, e_3$ correctly from the power sums. $e_1 = 151$ $e_2 = 5931$ $e_3 = 67221$ The integral $I$ was evaluated using the residue theorem. The pole is at $z_0 = \frac{205}{69}i$. Since the imaginary part is positive, the pole is inside the contour (assuming standard semi-circle in upper half plane). The residue calculation formula used was $\frac{1}{3!} \frac{d^3}{dz^3} [Q(z)e^{iz}]$. The code output $I \approx -3632.072 - 8.919i$. $\alpha \approx -3632.072$ $\beta \approx -8.919$ The quantity to compute is $\lfloor 10\alpha \rfloor + \lfloor \beta \rfloor$. $10\alpha \approx -36320.72 \implies \lfloor 10\alpha \rfloor = -36321$. $\beta \approx -8.919 \implies \lfloor \beta \rfloor = -9$. Sum = $-36321 + (-9) = -36330$. The code output ans_floor = -36330 confirms this. Wait, let me double check the int() vs floor() logic in the code output. The code printed ans_floor=-36330. math.floor(-36320.72) is -36321. math.floor(-8.919) is -9. Sum is -36330. The code output is consistent.

Final check on the integral formula. $\int_{-\infty}^\infty \frac{P(x)}{(x-z_0)^{n+1}} dx = 2\pi i \frac{1}{n!} \frac{d^n}{dz^n} [P(z)]_{z=z_0}$? No. The integrand is $f(x) = \frac{Q(x)e^{ix}}{(x-z_0)^4}$. Let $g(z) = Q(z)e^{iz}$. The residue at $z_0$ (order 4) is $\frac{1}{3!} g'''(z_0)$. So $I = 2\pi i \cdot \frac{1}{6} g'''(z_0)$. The code implemented exactly this. $g'''(z) = \frac{d^3}{dz^3} (Q(z)e^{iz})$. Using Leibniz rule or direct expansion: $g'(z) = Q' e^{iz} + i Q e^{iz} = (Q' + iQ) e^{iz}$ $g''(z) = (Q'' + iQ') e^{iz} + i(Q' + iQ) e^{iz} = (Q'' + 2iQ' - Q) e^{iz}$ $g'''(z) = (Q''' + 2iQ'' - Q') e^{iz} + i(Q'' + 2iQ' - Q) e^{iz} = (Q''' + 3iQ'' - 3Q' - iQ) e^{iz}$. The code used term_bracket = Q3 + 3j * Q2 - 3 * Q1 - 1j * Q0. This matches $(Q''' + 3iQ'' - 3Q' - iQ)$. The factor $e^{iz_0}$ was included. The factor $2\pi i / 6$ was included. The calculation seems correct.

One detail: $z_0 = \frac{205}{69}i$. $e^{i z_0} = e^{i (205/69) i} = e^{-205/69}$. This is a real number. $Q(z)$ coefficients are real. $z_0$ is purely imaginary. Let $z_0 = ki$. $Q(ki) = (ki)^3 - e_1(ki)^2 + e_2(ki) - e_3 = -k^3 i + e_1 k^2 + e_2 k i - e_3 = (e_1 k^2 - e_3) + i(e_2 k - k^3)$. $Q'(ki) = 3(ki)^2 - 2e_1(ki) + e_2 = -3k^2 - 2e_1 k i + e_2 = (e_2 - 3k^2) - i(2e_1 k)$. $Q''(ki) = 6ki - 2e_1 = -2e_1 + i(6k)$. $Q'''(ki) = 6$.

Term bracket $B = Q''' + 3iQ'' - 3Q' - iQ$. $B = 6 + 3i(-2e_1 + 6ki) - 3((e_2 - 3k^2) - 2ie_1 k) - i((e_1 k^2 - e_3) + i(e_2 k - k^3))$ $B = 6 - 6ie_1 - 18k - 3e_2 + 9k^2 + 6ie_1 k - ie_1 k^2 + ie_3 + e_2 k - k^3$ Real part of B: $6 - 18k - 3e_2 + 9k^2 + e_2 k - k^3$. Imaginary part of B: $-6e_1 + 6e_1 k - e_1 k^2 + e_3$.

Then $I = \frac{i \pi}{3} B e^{-k}$. $I = \frac{\pi}{3} e^{-k} (i \text{Re}(B) - \text{Im}(B))$. So $\text{Re}(I) = -\frac{\pi}{3} e^{-k} \text{Im}(B)$. $\text{Im}(I) = \frac{\pi}{3} e^{-k} \text{Re}(B)$.

The code output: $\alpha = \text{Re}(I) \approx -3632.07$. $\beta = \text{Im}(I) \approx -8.92$. The signs and magnitudes are handled by the code. The result ans_floor is -36330. [/THOUGHT]

\boxed{-36330}