Problem Analysis #2

[THOUGHT] Let $f(x) = (x-3i)^8$. Then we can expand this as

We are given $P(x) = x^7 - 14ix^6 + (85-60i)x^5 - (240+210i)x^4 + (420-840i)x^3 - (560+1680i)x^2 + (1344-2688i)x + 3840$. We observe that $P(x)$ is the expansion of $(x-6i)^7$. Specifically, we have

However, we are given $P(x) = x^7 - 14ix^6 + (85-60i)x^5 - (240+210i)x^4 + (420-840i)x^3 - (560+1680i)x^2 + (1344-2688i)x + 3840$. We can rewrite $P(x)$ as $(x-2i)^7 = x^7 - 14ix^6 + (85-60i)x^5 - (240+210i)x^4 + (420-840i)x^3 - (560+1680i)x^2 + (1344-2688i)x - 128i$. The given $P(x)$ is actually $(x-4i)^7 = x^7 - 28ix^6 + (336-112i)x^5 - (1344+896i) x^4 + (3584-2016i)x^3 - (5376+1792i) x^2 + (4032-504i)x - 16384 i$. Let us consider $(x-2i)^7 = x^7 - 14ix^6 + (85-60i)x^5 - (240+210i)x^4 + (420-840i)x^3 - (560+1680i)x^2 + (1344-2688i)x - 128i$. Then we have $P(x) = (x-4i)^7 + 3840$. We want to evaluate $\int_{-\infty}^{\infty} \frac{(x-2i)^7 e^{ix}}{(x-3i)^8}\,dx$. Using Cauchy's integral formula, we have $\int_{-\infty}^{\infty} \frac{P(x) e^{ix}}{(x-3i)^8}\,dx = 2\pi i f^{(7)}(3i)$. Let $f(x) = \frac{P(x)}{(x-3i)}$. Then we have $I = 2\pi i f^{(7)}(3i)$. Since $P(x) = (x-2i)^7$, we have $\int_{-\infty}^{\infty} \frac{(x-2i)^7 e^{ix}}{(x-3i)^8}\,dx$. We have $P(x) = (x-2i)^7$, so we want to evaluate $\int_{-\infty}^{\infty} \frac{(x-2i)^7 e^{ix}}{(x-3i)^8}\,dx$. Let $f(z) = (z-2i)^7$. Then we want to evaluate $\int_{-\infty}^{\infty} \frac{f(x) e^{ix}}{(x-3i)^8}\,dx$. We have $\int_{-\infty}^{\infty} \frac{P(x) e^{ix}}{(x-3i)^8}\,dx = 2\pi i \frac{d^7}{dz^7} P(z) |_{z=3i}$. Since $P(x)=(x-2i)^7$, we have $\frac{d^7}{dx^7} P(x) = 7!$. Then $\int_{-\infty}^{\infty} \frac{(x-2i)^7 e^{ix}}{(x-3i)^8}\,dx = 2\pi i (7!) |_{z=3i}$. However, we are given $P(x) = x^7 - 14ix^6 + (85-60i)x^5 - (240+210i)x^4 + (420-840i)x^3 - (560+1680i)x^2 + (1344-2688i)x + 3840$. We have $P(x)=(x-2i)^7 + 3840$. Then we have $\int_{-\infty}^{\infty} \frac{((x-2i)^7 + 3840) e^{ix}}{(x-3i)^8}\,dx$. Let $f(z) = (z-2i)^7 + 3840$. Then $f^{(7)}(z) = 7!$ for all z. Then $\int_{-\infty}^{\infty} \frac{((x-2i)^7 + 3840) e^{ix}}{(x-3i)^8}\,dx = 2\pi i (7!) |_{z=3i} = 2\pi i (5040)$. Then $I = \int_{-\infty}^{\infty} \frac{P(x) e^{ix}}{(x-3i)^8}\,dx = 2\pi i \frac{d^7}{dz^7} ((z-2i)^7 + 3840) |_{z=3i}$. Since $\frac{d^7}{dz^7} ((z-2i)^7 + 3840) = 7!$, we have $I = 2\pi i (5040) = 10080 \pi i$. Then $a=0$ and $b = 10080 \pi$, so $\lfloor 10a \rfloor + \lfloor b \rfloor = \lfloor 0 \rfloor + \lfloor 10080\pi \rfloor = 0 + 31541 = 31541$.

[PYTHON] import numpy as np

a = 0.0 b = 10080 * np.pi print(int(np.floor(10*a)) + int(np.floor(b)))

[THOUGHT] The Python output gives 31667.

$\boxed{31667}$