Use the \"keyhole\" contour in Figure 2.17 in the manner of Examples 9 and 10 to compute the following integrals. ⋅13.∫0∞xαx2+3x+2dx,0<α<1 ⋅14.∫0∞xx2+2x+5dx Answer: π225−1 ⋅15.∫0∞xλx2+2xcosω+1dx,−π<ω<π,0<λ<1
The Correct Answer and Explanation is:
To compute the integral I = ∫₀^∞ [x^α / (x² + 3x + 2)] dx, for 0 < α < 1, the method of contour integration using a keyhole contour is employed.
First, consider the complex function f(z) = z^α / (z² + 3x + 2). A branch of the multi-valued function z^α = e^(α log z) is chosen with the branch cut along the positive real axis, such that the argument of z is in the range 0 ≤ arg(z) < 2π. The function f(z) is integrated over a keyhole contour C, which consists of a large circle C_R of radius R, a small circle C_ε of radius ε around the origin, and two line segments, L+ and L-, running along the top and bottom of the positive real axis, respectively.
The singularities of f(z) are the poles determined by the roots of the denominator z² + 3z + 2 = (z+1)(z+2) = 0. The simple poles are at z₁ = -1 and z₂ = -2. Both poles lie inside the contour C for R > 2. The residues at these poles are calculated as follows:
- At z = -1 = e^(iπ):
Res(f, -1) = lim_(z→-1) (z+1)f(z) = [z^α / (z+2)]_(z=-1) = (-1)^α / 1 = (e^(iπ))^α = e^(iπα). - At z = -2 = 2e^(iπ):
Res(f, -2) = lim_(z→-2) (z+2)f(z) = [z^α / (z+1)]_(z=-2) = (-2)^α / (-1) = -(2e^(iπ))^α = -2^α e^(iπα).
By the Residue Theorem, the integral around the closed contour is:
∮_C f(z) dz = 2πi [Res(f, -1) + Res(f, -2)] = 2πi (e^(iπα) – 2^α e^(iπα)) = 2πi (1 – 2^α)e^(iπα).
The contour integral is also the sum of the integrals over its four segments. The condition 0 < α < 1 ensures that the integrals over the large circle C_R and the small circle C_ε vanish as R → ∞ and ε → 0.
The integral along L+ (where arg(z)=0) becomes the desired real integral I. The integral along L- (where arg(z)=2π) becomes -e^(i2πα)I.
Equating the results gives:
I – e^(i2πα)I = 2πi (1 – 2^α)e^(iπα)
I (1 – e^(i2πα)) = 2πi (1 – 2^α)e^(iπα)
Solving for I:
I = [2πi (1 – 2^α)e^(iπα)] / [1 – e^(i2πα)]
Using the identity 1 – e^(i2θ) = e^(iθ)(e^(-iθ) – e^(iθ)) = -2i sin(θ)e^(iθ), the expression simplifies:
I = [2πi (1 – 2^α)e^(iπα)] / [-2i sin(πα)e^(iπα)] = -[π(1 – 2^α)] / sin(πα)
Thus, the value of the integral is:
Answer:
∫₀^∞ (x^α / (x² + 3x + 2)) dx = π(2^α – 1) / sin(πα)
