[THOUGHT]
The given integral involves a polynomial P ( x ) P(x) P ( x ) e i x e^{ix} e i x ( x − 3 i ) 8 (x - 3i)^8 ( x − 3 i ) 8 x = 3 i x = 3i x = 3 i
Here, the integrand P ( x ) e i x ( x − 3 i ) 8 \frac{P(x) e^{ix}}{(x-3i)^8} ( x − 3 i ) 8 P ( x ) e i x x = 3 i x = 3i x = 3 i x = 3 i x = 3i x = 3 i n n n
Res ( P ( x ) e i x ( x − 3 i ) 8 , 3 i ) = 1 ( 8 − 1 ) ! d 7 d x 7 [ ( x − 3 i ) 8 ⋅ P ( x ) e i x ( x − 3 i ) 8 ] ∣ x = 3 i = 1 7 ! d 7 d x 7 [ P ( x ) e i x ] ∣ x = 3 i . \text{Res}\left(\frac{P(x) e^{ix}}{(x-3i)^8}, 3i\right) = \frac{1}{(8-1)!} \left. \frac{d^{7}}{dx^{7}} \left[ (x-3i)^8 \cdot \frac{P(x) e^{ix}}{(x-3i)^8} \right] \right|_{x=3i} = \frac{1}{7!} \left. \frac{d^{7}}{dx^{7}} \left[ P(x) e^{ix} \right] \right|_{x=3i}. Res ( ( x − 3 i ) 8 P ( x ) e i x , 3 i ) = ( 8 − 1 )! 1 d x 7 d 7 [ ( x − 3 i ) 8 ⋅ ( x − 3 i ) 8 P ( x ) e i x ] x = 3 i = 7 ! 1 d x 7 d 7 [ P ( x ) e i x ] x = 3 i . Let's compute the 7th derivative of P ( x ) e i x P(x) e^{ix} P ( x ) e i x
We have:
P ( x ) = x 7 − 14 i x 6 + ( 85 − 60 i ) x 5 − ( 240 + 210 i ) x 4 + ( 420 − 840 i ) x 3 − ( 560 + 1680 i ) x 2 + ( 1344 − 2688 i ) 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. P ( x ) = x 7 − 14 i x 6 + ( 85 − 60 i ) x 5 − ( 240 + 210 i ) x 4 + ( 420 − 840 i ) x 3 − ( 560 + 1680 i ) x 2 + ( 1344 − 2688 i ) x + 3840. The 7th derivative of P ( x ) e i x P(x) e^{ix} P ( x ) e i x
d 7 d x 7 [ P ( x ) e i x ] = ∑ k = 0 7 ( 7 k ) P ( k ) ( x ) ( e i x ) ( 7 − k ) . \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right] = \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(x) \left( e^{ix} \right)^{(7-k)}. d x 7 d 7 [ P ( x ) e i x ] = k = 0 ∑ 7 ( k 7 ) P ( k ) ( x ) ( e i x ) ( 7 − k ) . Calculating the derivatives of P ( x ) P(x) P ( x )
P ′ ( x ) = 7 x 6 − 84 i x 5 + 5 ( 85 − 60 i ) x 4 − 4 ( 240 + 210 i ) x 3 + 3 ( 420 − 840 i ) x 2 − 2 ( 560 + 1680 i ) x + ( 1344 − 2688 i ) , P'(x) = 7x^6 - 84ix^5 + 5(85-60i)x^4 - 4(240+210i)x^3 + 3(420-840i)x^2 - 2(560+1680i)x + (1344-2688i), P ′ ( x ) = 7 x 6 − 84 i x 5 + 5 ( 85 − 60 i ) x 4 − 4 ( 240 + 210 i ) x 3 + 3 ( 420 − 840 i ) x 2 − 2 ( 560 + 1680 i ) x + ( 1344 − 2688 i ) , P ′ ′ ( x ) = 42 x 5 − 420 i x 4 + 20 ( 85 − 60 i ) x 3 − 12 ( 240 + 210 i ) x 2 + 6 ( 420 − 840 i ) x − 2 ( 560 + 1680 i ) , P''(x) = 42x^5 - 420ix^4 + 20(85-60i)x^3 - 12(240+210i)x^2 + 6(420-840i)x - 2(560+1680i), P ′′ ( x ) = 42 x 5 − 420 i x 4 + 20 ( 85 − 60 i ) x 3 − 12 ( 240 + 210 i ) x 2 + 6 ( 420 − 840 i ) x − 2 ( 560 + 1680 i ) , P ′ ′ ′ ( x ) = 210 x 4 − 1680 i x 3 + 60 ( 85 − 60 i ) x 2 − 24 ( 240 + 210 i ) x + 6 ( 420 − 840 i ) , P'''(x) = 210x^4 - 1680ix^3 + 60(85-60i)x^2 - 24(240+210i)x + 6(420-840i), P ′′′ ( x ) = 210 x 4 − 1680 i x 3 + 60 ( 85 − 60 i ) x 2 − 24 ( 240 + 210 i ) x + 6 ( 420 − 840 i ) , P ( 4 ) ( x ) = 840 x 3 − 5040 i x 2 + 120 ( 85 − 60 i ) x − 24 ( 240 + 210 i ) , P^{(4)}(x) = 840x^3 - 5040ix^2 + 120(85-60i)x - 24(240+210i), P ( 4 ) ( x ) = 840 x 3 − 5040 i x 2 + 120 ( 85 − 60 i ) x − 24 ( 240 + 210 i ) , P ( 5 ) ( x ) = 2520 x 2 − 10080 i x + 120 ( 85 − 60 i ) , P^{(5)}(x) = 2520x^2 - 10080ix + 120(85-60i), P ( 5 ) ( x ) = 2520 x 2 − 10080 i x + 120 ( 85 − 60 i ) , P ( 6 ) ( x ) = 5040 x − 10080 i , P^{(6)}(x) = 5040x - 10080i, P ( 6 ) ( x ) = 5040 x − 10080 i , P ( 7 ) ( x ) = 5040. P^{(7)}(x) = 5040. P ( 7 ) ( x ) = 5040. Now, calculating the derivatives of e i x e^{ix} e i x
( e i x ) ′ = i e i x , ( e i x ) ′ ′ = − e i x , ( e i x ) ′ ′ ′ = − i e i x , ( e i x ) ( 4 ) = e i x , ( e i x ) ( 5 ) = i e i x , ( e i x ) ( 6 ) = − e i x , ( e i x ) ( 7 ) = − i e i x . \left( e^{ix} \right)' = ie^{ix}, \quad \left( e^{ix} \right)'' = -e^{ix}, \quad \left( e^{ix} \right)''' = -ie^{ix}, \quad \left( e^{ix} \right)^{(4)} = e^{ix}, \quad \left( e^{ix} \right)^{(5)} = ie^{ix}, \quad \left( e^{ix} \right)^{(6)} = -e^{ix}, \quad \left( e^{ix} \right)^{(7)} = -ie^{ix}. ( e i x ) ′ = i e i x , ( e i x ) ′′ = − e i x , ( e i x ) ′′′ = − i e i x , ( e i x ) ( 4 ) = e i x , ( e i x ) ( 5 ) = i e i x , ( e i x ) ( 6 ) = − e i x , ( e i x ) ( 7 ) = − i e i x . Substituting x = 3 i x = 3i x = 3 i
d 7 d x 7 [ P ( x ) e i x ] ∣ x = 3 i = 5040 ( − i e i ( 3 i ) ) + 7 ⋅ 5040 ( − e i ( 3 i ) ) + 21 ⋅ 10080 i ( e i ( 3 i ) ) + 35 ⋅ 120 ( 85 − 60 i ) ( − i e i ( 3 i ) ) + 35 ⋅ 120 ( 240 + 210 i ) ( − e i ( 3 i ) ) + 21 ⋅ 5040 ( − i e i ( 3 i ) ) + 7 ⋅ 840 ( e i ( 3 i ) ) + 42 ( − i e i ( 3 i ) ) . \left. \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right] \right|_{x=3i} = 5040(-ie^{i(3i)}) + 7 \cdot 5040(-e^{i(3i)}) + 21 \cdot 10080i(e^{i(3i)}) + 35 \cdot 120(85-60i)(-ie^{i(3i)}) + 35 \cdot 120(240+210i)(-e^{i(3i)}) + 21 \cdot 5040(-ie^{i(3i)}) + 7 \cdot 840(e^{i(3i)}) + 42(-ie^{i(3i)}). d x 7 d 7 [ P ( x ) e i x ] x = 3 i = 5040 ( − i e i ( 3 i ) ) + 7 ⋅ 5040 ( − e i ( 3 i ) ) + 21 ⋅ 10080 i ( e i ( 3 i ) ) + 35 ⋅ 120 ( 85 − 60 i ) ( − i e i ( 3 i ) ) + 35 ⋅ 120 ( 240 + 210 i ) ( − e i ( 3 i ) ) + 21 ⋅ 5040 ( − i e i ( 3 i ) ) + 7 ⋅ 840 ( e i ( 3 i ) ) + 42 ( − i e i ( 3 i ) ) . First, calculate e i ( 3 i ) = e − 3 e^{i(3i)} = e^{-3} e i ( 3 i ) = e − 3
e i ( 3 i ) = e − 3 . e^{i(3i)} = e^{-3}. e i ( 3 i ) = e − 3 . Substitute e − 3 e^{-3} e − 3
d 7 d x 7 [ P ( x ) e i x ] ∣ x = 3 i = 5040 ( − i e − 3 ) + 7 ⋅ 5040 ( − e − 3 ) + 21 ⋅ 10080 i ( e − 3 ) + 35 ⋅ 120 ( 85 − 60 i ) ( − i e − 3 ) + 35 ⋅ 120 ( 240 + 210 i ) ( − e − 3 ) + 21 ⋅ 5040 ( − i e − 3 ) + 7 ⋅ 840 ( e − 3 ) + 42 ( − i e − 3 ) . \left. \frac{d^7}{dx^7} \left[ P(x) e^{ix} \right] \right|_{x=3i} = 5040(-i e^{-3}) + 7 \cdot 5040(-e^{-3}) + 21 \cdot 10080i(e^{-3}) + 35 \cdot 120(85-60i)(-ie^{-3}) + 35 \cdot 120(240+210i)(-e^{-3}) + 21 \cdot 5040(-ie^{-3}) + 7 \cdot 840(e^{-3}) + 42(-ie^{-3}). d x 7 d 7 [ P ( x ) e i x ] x = 3 i = 5040 ( − i e − 3 ) + 7 ⋅ 5040 ( − e − 3 ) + 21 ⋅ 10080 i ( e − 3 ) + 35 ⋅ 120 ( 85 − 60 i ) ( − i e − 3 ) + 35 ⋅ 120 ( 240 + 210 i ) ( − e − 3 ) + 21 ⋅ 5040 ( − i e − 3 ) + 7 ⋅ 840 ( e − 3 ) + 42 ( − i e − 3 ) . Simplify:
= − 5040 i e − 3 − 35280 e − 3 + 210576 i e − 3 − 35 ⋅ 120 ⋅ 85 i e − 3 + 35 ⋅ 120 ⋅ 60 e − 3 − 35 ⋅ 120 ⋅ 240 e − 3 − 35 ⋅ 120 ⋅ 210 i e − 3 − 105680 i e − 3 + 5880 e − 3 − 42 i e − 3 . = -5040ie^{-3} - 35280e^{-3} + 210576ie^{-3} - 35 \cdot 120 \cdot 85ie^{-3} + 35 \cdot 120 \cdot 60 e^{-3} - 35 \cdot 120 \cdot 240 e^{-3} - 35 \cdot 120 \cdot 210 i e^{-3} - 105680ie^{-3} + 5880e^{-3} - 42ie^{-3}. = − 5040 i e − 3 − 35280 e − 3 + 210576 i e − 3 − 35 ⋅ 120 ⋅ 85 i e − 3 + 35 ⋅ 120 ⋅ 60 e − 3 − 35 ⋅ 120 ⋅ 240 e − 3 − 35 ⋅ 120 ⋅ 210 i e − 3 − 105680 i e − 3 + 5880 e − 3 − 42 i e − 3 . Combine like terms:
= ( − 5040 i + 210576 i − 35 ⋅ 120 ⋅ 85 i − 35 ⋅ 120 ⋅ 210 i − 105680 i − 42 i ) e − 3 + ( − 35280 + 35 ⋅ 120 ⋅ 60 − 35 ⋅ 120 ⋅ 240 + 5880 ) e − 3 . = (-5040i + 210576i - 35 \cdot 120 \cdot 85i - 35 \cdot 120 \cdot 210i - 105680i - 42i)e^{-3} + (-35280 + 35 \cdot 120 \cdot 60 - 35 \cdot 120 \cdot 240 + 5880)e^{-3}. = ( − 5040 i + 210576 i − 35 ⋅ 120 ⋅ 85 i − 35 ⋅ 120 ⋅ 210 i − 105680 i − 42 i ) e − 3 + ( − 35280 + 35 ⋅ 120 ⋅ 60 − 35 ⋅ 120 ⋅ 240 + 5880 ) e − 3 . Calculate the coefficients:
− 5040 i + 210576 i − 35 ⋅ 120 ⋅ 85 i − 35 ⋅ 120 ⋅ 210 i − 105680 i − 42 i = − 5040 i + 210576 i − 35700 i − 85800 i − 105680 i − 42 i = − 42 ⋅ 120 ⋅ 210 i = − 10582 i , -5040i + 210576i - 35 \cdot 120 \cdot 85i - 35 \cdot 120 \cdot 210i - 105680i - 42i = -5040i + 210576i - 35700i - 85800i - 105680i - 42i = -42 \cdot 120 \cdot 210i = -10582i, − 5040 i + 210576 i − 35 ⋅ 120 ⋅ 85 i − 35 ⋅ 120 ⋅ 210 i − 105680 i − 42 i = − 5040 i + 210576 i − 35700 i − 85800 i − 105680 i − 42 i = − 42 ⋅ 120 ⋅ 210 i = − 10582 i , − 35280 + 35 ⋅ 120 ⋅ 60 − 35 ⋅ 120 ⋅ 240 + 5880 = − 35280 + 25200 − 100800 + 5880 = − 10200. -35280 + 35 \cdot 120 \cdot 60 - 35 \cdot 120 \cdot 240 + 5880 = -35280 + 25200 - 100800 + 5880 = -10200. − 35280 + 35 ⋅ 120 ⋅ 60 − 35 ⋅ 120 ⋅ 240 + 5880 = − 35280 + 25200 − 100800 + 5880 = − 10200. Thus, the residue is:
Res = 1 7 ! ( − 10582 i e − 3 − 10200 e − 3 ) = − 10582 i − 10200 5040 e 3 . \text{Res} = \frac{1}{7!} (-10582ie^{-3} - 10200e^{-3}) = \frac{-10582i - 10200}{5040e^3}. Res = 7 ! 1 ( − 10582 i e − 3 − 10200 e − 3 ) = 5040 e 3 − 10582 i − 10200 . The integral is 2 π i 2\pi i 2 πi
I = 2 π i ⋅ − 10582 i − 10200 5040 e 3 = 2 π ⋅ − 10582 i 2 − 10200 i 5040 e 3 = 2 π ⋅ 10582 − 10200 i 5040 e 3 = 2 π ( 10582 − 10200 i ) 5040 e 3 . I = 2\pi i \cdot \frac{-10582i - 10200}{5040e^3} = 2\pi \cdot \frac{-10582i^2 - 10200i}{5040e^3} = 2\pi \cdot \frac{10582 - 10200i}{5040e^3} = \frac{2\pi (10582 - 10200i)}{5040e^3}. I = 2 πi ⋅ 5040 e 3 − 10582 i − 10200 = 2 π ⋅ 5040 e 3 − 10582 i 2 − 10200 i = 2 π ⋅ 5040 e 3 10582 − 10200 i = 5040 e 3 2 π ( 10582 − 10200 i ) . Simplify:
I = π ( 10582 − 10200 i ) 2520 e 3 = 10582 π 2520 e 3 − 10200 π i 2520 e 3 = 10582 π 2520 e 3 − 5 π i e 3 . I = \frac{\pi (10582 - 10200i)}{2520e^3} = \frac{10582\pi}{2520e^3} - \frac{10200\pi i}{2520e^3} = \frac{10582\pi}{2520e^3} - \frac{5\pi i}{e^3}. I = 2520 e 3 π ( 10582 − 10200 i ) = 2520 e 3 10582 π − 2520 e 3 10200 πi = 2520 e 3 10582 π − e 3 5 πi . Calculate the real and imaginary parts:
Using Python, compute the numerical values of 10582 π 2520 e 3 \frac{10582\pi}{2520e^3} 2520 e 3 10582 π − 5 π e 3 -\frac{5\pi}{e^3} − e 3 5 π
Constants
pi = math.pi
e = math.e
Real part
real_part = (10582 * pi) / (2520 * e**3)
Imaginary part
imaginary_part = -(5 * pi) / e**3
print(f"Real part: {real_part}")
print(f"Imaginary part: {imaginary_part}")