Answer on Question #50254 – Math – Complex Analysis
1. Use Residue theorem to compute
Principle value
integral from ( - infinity to infinity )
[ 3 x ] / [ ( x 2 + 1 ) 2 ( x + 2 ) ] dx [3x] / [(x^2 + 1)^{2}(x + 2)] \text{dx} [ 3 x ] / [( x 2 + 1 ) 2 ( x + 2 )] dx
Solution.
I = ∫ − ∞ + ∞ 3 x d x ( x 2 + 1 ) 2 ( x + 2 ) = ∫ C + f ( z ) d z − ∫ z = − 2 z = 0 f ( z ) d z , where f ( z ) = 3 z ( z 2 + 1 ) 2 ( z + 2 ) . I = \int_{-\infty}^{+\infty} \frac{3x\,dx}{(x^2 + 1)^2(x + 2)} = \int_{C^+} f(z)\,dz - \int_{z=-2}^{z=0} f(z)\,dz, \quad \text{where} \quad f(z) = \frac{3z}{(z^2 + 1)^2(z + 2)}. I = ∫ − ∞ + ∞ ( x 2 + 1 ) 2 ( x + 2 ) 3 x d x = ∫ C + f ( z ) d z − ∫ z = − 2 z = 0 f ( z ) d z , where f ( z ) = ( z 2 + 1 ) 2 ( z + 2 ) 3 z .
The isolated singular points of the function are z = − 2 z = -2 z = − 2 z = ± i z = \pm i z = ± i z = i z = i z = i X X X res z = i f = lim z → i d d z f ⋅ ( z − i ) 2 = \text{res}_{z=i} f = \lim_{z \to i} \frac{d}{dz} f \cdot (z - i)^2 = res z = i f = lim z → i d z d f ⋅ ( z − i ) 2 =
= lim z → i d d z 3 z ( z + i ) 2 ( z + 2 ) = 3 lim z → i 1 ⋅ ( z + i ) 2 ( z + 2 ) − z ⋅ [ 2 ( z + i ) ( z + 2 ) + ( z + i ) 2 ] ( z + i ) 4 ( z + 2 ) 2 = = lim z → i 6 ( i − z 2 − z ) ( z + i ) 3 ( z + 2 ) 2 = 3 i 4 ( i + 2 ) 2 . \begin{aligned}
&= \lim_{z \to i} \frac{d}{dz} \frac{3z}{(z + i)^2(z + 2)} = 3 \lim_{z \to i} \frac{1 \cdot (z + i)^2(z + 2) - z \cdot [2(z + i)(z + 2) + (z + i)^2]}{(z + i)^4(z + 2)^2} = \\
&= \lim_{z \to i} \frac{6(i - z^2 - z)}{(z + i)^3(z + 2)^2} = \frac{3i}{4(i + 2)^2}.
\end{aligned} = z → i lim d z d ( z + i ) 2 ( z + 2 ) 3 z = 3 z → i lim ( z + i ) 4 ( z + 2 ) 2 1 ⋅ ( z + i ) 2 ( z + 2 ) − z ⋅ [ 2 ( z + i ) ( z + 2 ) + ( z + i ) 2 ] = = z → i lim ( z + i ) 3 ( z + 2 ) 2 6 ( i − z 2 − z ) = 4 ( i + 2 ) 2 3 i . ∫ z = − 2 z = 0 f ( z ) d z = ∣ z = − 2 + r e i φ d z = r e i φ ∣ = lim r → + 0 ∫ π 0 3 ( − 2 + r e i φ ) r e i φ i d φ ( ( − 2 + r e i φ ) 2 + 1 ) 2 r e i φ = ∫ π 0 3 ( − 2 ) i d φ ( 2 2 + 1 ) 2 = − 6 i 25 ∫ π 0 d φ = 6 i π 25 . \int_{z=-2}^{z=0} f(z)\,dz = \left| \begin{array}{l} z = -2 + r e^{i\varphi} \\ dz = r e^{i\varphi} \end{array} \right| = \lim_{r \to +0} \int_{\pi}^{0} \frac{3(-2 + r e^{i\varphi}) r e^{i\varphi} \, i\,d\varphi}{\left( (-2 + r e^{i\varphi})^2 + 1 \right)^2 r e^{i\varphi}} = \int_{\pi}^{0} \frac{3(-2) i d\varphi}{(2^2 + 1)^2} = -\frac{6i}{25} \int_{\pi}^{0} d\varphi = \frac{6i\pi}{25}. ∫ z = − 2 z = 0 f ( z ) d z = ∣ ∣ z = − 2 + r e i φ d z = r e i φ ∣ ∣ = r → + 0 lim ∫ π 0 ( ( − 2 + r e i φ ) 2 + 1 ) 2 r e i φ 3 ( − 2 + r e i φ ) r e i φ i d φ = ∫ π 0 ( 2 2 + 1 ) 2 3 ( − 2 ) i d φ = − 25 6 i ∫ π 0 d φ = 25 6 iπ .
According to the residue theorem, I = 2 π i ⋅ res z = i f − 6 i π 25 = 2 π i ⋅ 3 i 4 ( i + 2 ) 2 − 6 i π 25 = − 9 π 50 I = 2\pi i \cdot \text{res}_{z=i} f - \frac{6i\pi}{25} = 2\pi i \cdot \frac{3i}{4(i + 2)^2} - \frac{6i\pi}{25} = -\frac{9\pi}{50} I = 2 πi ⋅ res z = i f − 25 6 iπ = 2 πi ⋅ 4 ( i + 2 ) 2 3 i − 25 6 iπ = − 50 9 π
Answer: − 9 π 50 -\frac{9\pi}{50} − 50 9 π
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