Problem Analysis #2

Deepseek-V3.2 · Run 1

135.38s
4,821 toks
Problem Statement

Evaluate the integral

I=P(x)eix(x3i)8dx,I = \int_{-\infty}^{\infty} \frac{P(x)\, e^{ix}}{(x - 3i)^{8}}\,dx,

where

P(x)=x714ix6+(8560i)x5(240+210i)x4+(420840i)x3(560+1680i)x2+(13442688i)x+3840.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.

If the value of the integral is I=a+biI = a + bi, ii is the imaginary unit, compute

10a+b.\lfloor 10a \rfloor + \lfloor b \rfloor.
Ground Truth
7
Execution Trace
Reasoning Process

[THOUGHT] We need to evaluate the integral

I=P(x)eix(x3i)8dxI = \int_{-\infty}^{\infty} \frac{P(x) e^{ix}}{(x - 3i)^{8}} dx

where P(x)P(x) is a polynomial of degree 7. The integrand is analytic except at x=3ix = 3i (pole of order 8). Since the integration path is along the real axis, we can close the contour in the upper half-plane if the integrand decays sufficiently for large x|x| in the upper half-plane. The factor eixe^{ix} decays in the upper half-plane because eix=ei(x+iy)=eixye^{ix} = e^{i(x+iy)} = e^{ix - y} with y>0y>0. So we can close the contour upward. Then by the residue theorem, I=2πiResidue at x=3iI = 2\pi i \cdot \text{Residue at } x = 3i (since the pole is inside the contour? Wait: x=3ix = 3i is in the upper half-plane (positive imaginary part), yes, because 3i3i has imaginary part 3 > 0. So indeed the pole is inside the closed contour (upper half-plane). So

I=2πiResx=3iP(x)eix(x3i)8.I = 2\pi i \cdot \text{Res}_{x=3i} \frac{P(x) e^{ix}}{(x - 3i)^{8}}.

Since the denominator is (x3i)8(x-3i)^8, the residue for a pole of order nn is given by

Resz=z0f(z)=1(n1)!limzz0dn1dzn1[(zz0)nf(z)].\text{Res}_{z=z_0} f(z) = \frac{1}{(n-1)!} \lim_{z\to z_0} \frac{d^{n-1}}{dz^{n-1}} \left[ (z-z_0)^n f(z) \right].

Here n=8n=8, so

Resx=3i=17!limx3id7dx7[(x3i)8P(x)eix(x3i)8]=17!limx3id7dx7[P(x)eix].\text{Res}_{x=3i} = \frac{1}{7!} \lim_{x\to 3i} \frac{d^7}{dx^7} \left[ (x-3i)^8 \cdot \frac{P(x) e^{ix}}{(x-3i)^{8}} \right] = \frac{1}{7!} \lim_{x\to 3i} \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right].

Thus

I=2πi17!d7dx7[P(x)eix]x=3i.I = 2\pi i \cdot \frac{1}{7!} \left. \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right] \right|_{x=3i}.

We need to compute the 7th derivative of P(x)eixP(x) e^{ix} evaluated at x=3ix=3i. Since P(x)P(x) is a polynomial of degree 7, its 7th derivative will be constant (the leading term's derivative). Let's compute.

Let f(x)=P(x)eixf(x) = P(x) e^{ix}. Use Leibniz rule:

d7dx7[P(x)eix]=k=07(7k)P(k)(x)(eix)(7k).\frac{d^7}{dx^7} [P(x) e^{ix}] = \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(x) \cdot (e^{ix})^{(7-k)}.

But (eix)(m)=imeix(e^{ix})^{(m)} = i^m e^{ix}. So

=k=07(7k)P(k)(x)i7keix.= \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(x) i^{7-k} e^{ix}.

Evaluate at x=3ix=3i.

Alternatively, note that P(x)P(x) is degree 7, so P(7)(x)=7!P^{(7)}(x) = 7! \cdot leading coefficient. The leading term of P(x)P(x) is x7x^7 (coefficient 1). So P(7)(x)=7!P^{(7)}(x) = 7!. Lower derivatives depend on polynomial coefficients.

We can compute explicitly using Python symbolic differentiation.

We'll compute P(x)P(x) as given, then compute the 7th derivative of P(x)eixP(x) e^{ix} symbolically and evaluate at x=3ix=3i. Then multiply by 2πi/7!2\pi i / 7! to get II. Then extract real and imaginary parts aa and bb, compute floor(10a) and floor(b), and sum them.

Let's do it step by step in Python. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

x = sp.symbols('x') i = sp.I # imaginary unit

Define polynomial P(x)

P = (x7 - 14ix6 + (85 - 60*i)x**5 - (240 + 210i)x**4 + (420 - 840i)x**3 - (560 + 1680i)x**2 + (1344 - 2688i)*x + 3840)

Define f(x) = P(x) * exp(i*x)

f = P * sp.exp(i*x)

Compute 7th derivative

f7 = sp.diff(f, x, 7)

Evaluate at x = 3i

val = sp.simplify(f7.subs(x, 3*i)) print("7th derivative at 3i:", val)

Compute residue: val / 7!

residue = val / sp.factorial(7) print("Residue:", residue)

Compute integral I = 2pii * residue

I = 2sp.pii * residue print("Integral I:", I) print("I simplified:", sp.simplify(I))

Extract real and imaginary parts

a = sp.re(I) b = sp.im(I) print("a =", a) print("b =", b)

Compute floor(10a) and floor(b)

floor_10a = sp.floor(10*a) floor_b = sp.floor(b) print("floor(10a) =", floor_10a) print("floor(b) =", floor_b)

Compute sum

result = floor_10a + floor_b print("Result:", result) [/PYTHON]

Tool Output
[PYTHON OUTPUT] 7th derivative at 3i: 6*(5479 - 1608*I)*exp(-3) Residue: (5479 - 1608*I)*exp(-3)/840 Integral I: I*pi*(5479 - 1608*I)*exp(-3)/420 I simplified: I*pi*(5479 - 1608*I)*exp(-3)/420 a = 134*pi*exp(-3)/35 b = 5479*pi*exp(-3)/420 floor(10a) = 5 floor(b) = 2 Result: 7 None [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] From the Python output, we have:

  • a=134πe335a = \frac{134\pi e^{-3}}{35}
  • b=5479πe3420b = \frac{5479\pi e^{-3}}{420}
  • 10a=5\lfloor 10a \rfloor = 5
  • b=2\lfloor b \rfloor = 2
  • Result = 5+2=75 + 2 = 7

Thus the required integer is 7. [/THOUGHT] 7\boxed{7}

Final Model Output
7

Comparative Analysis (P2) (Average across runs)

Avg Time Taken

Qwen3-30B-A3B-2507
7431.3
Deepseek-R1-distill-Qwen-32B (high)
859.1
Solar-Open-100B
426.8
Gemini-3-Pro-Preview
199.4
Deepseek-V3.2
116.7
Kanana-2-30B-Thinking-2601
92.2
GPT-oss-20B (high)
86.7
Gemma-3-27B
83.2
Kanana-2-30B-Thinking
72.1
Solar-Pro-2 (31B)(high)
71.2
HCX-007(high)
61.9
A.X-4.0 (72B)
42.6
Llama-VARCO-8B-Instruct
35.1
Grok-4.1-fast
22.6
K-EXAONE-236B-A23B
19.8
Claude-Opus-4.5
15.8
GPT-5.1 (high)
13.1
EXAONE-4.0.1-32B (high)
5.4

Avg Token Usage

Qwen3-30B-A3B-2507
24,110
GPT-oss-20B (high)
19,053
Kanana-2-30B-Thinking-2601
18,048
Gemini-3-Pro-Preview
17,645
Deepseek-R1-distill-Qwen-32B (high)
11,661
Kanana-2-30B-Thinking
11,648
HCX-007(high)
11,315
Solar-Pro-2 (31B)(high)
10,817
Solar-Open-100B
6,687
Deepseek-V3.2
5,996
A.X-4.0 (72B)
5,881
Grok-4.1-fast
5,630
K-EXAONE-236B-A23B
5,505
Llama-VARCO-8B-Instruct
5,237
Claude-Opus-4.5
4,039
EXAONE-4.0.1-32B (high)
3,955
Gemma-3-27B
3,895
GPT-5.1 (high)
3,378