复变函数习题解答(第6章)

p269第六章习题(一) [ 7, 8, 9, 10, 11, 12, 13, 14 ]

7. 从⎰C e i z /√z dz 出发，其中C 是如图所示之周线(√z 沿正实轴取正值)，证明：

⎰(0, +∞) cos x /√x dx = ⎰(0, +∞) sin x /√x dx = √(π/2)．

【解】| ⎰C (R ) e i z /√z dz | ≤ ⎰C (R ) | e i z |/R 1/2 ds

= ⎰[0, π/2] | e i ρ (cos θ + i sin θ )|/R 1/2 · R d θ = ⎰[0, π/2] | e - R sin θ | R 1/2 d θ

≤ R 1/2 ⎰[0, π/2] e - R sin θ d θ．

由sin θ ≥ 2θ/π (θ∈[0, π/2] )，故

R 1/2⎰[0, π/2] e - R sin θ d θ

≤ R 1/2 ⎰[0, π/2] e - (2R / π)θ d θ = (π/(2R 1/2))(1 – e - R ) ≤ π/(2R 1/2)．

所以，| ⎰C (R ) e i z /√z dz | → 0 (as R →+∞)．

而由| ⎰C (r ) e i z /√z dz | ≤ (π/(2r 1/2))(1 – e - r )

知| ⎰C (r ) e i z /√z dz | → 0 (as r → 0+ )．

当r → 0+，R →+∞时，

⎰[r , R ] e i z /√z dz = ⎰[r , R ] e i x /√x dx = ⎰[r , R ] (cos x + i sin x )/√x dx

→ ⎰(0, +∞) cos x /√x dx + i ⎰(0, +∞) sin x /√x dx ．

⎰[r i , R i ] e i z /√z dz = ⎰[r , R ] e i (i y )/√(i y ) i dy = ⎰[r , R ] e - y e i π/4/√y dy ．

= (1 + i )/√2 · ⎰[r , R ] e - y /√y dy = 2(1 + i )/√2 · ⎰[√r , √R ] e - u ^2 du

→ (1 + i )√2 · ⎰(0, +∞) e - u ^2 du = (1 + i )√2 · √π/2 = (1 + i )√(π/2)．

由Cauchy 积分定理，⎰C e i z /√z dz = 0，故其极限也为0，

所以，⎰(0, +∞) cos x /√x dx + i ⎰(0, +∞) sin x /√x dx = (1 + i )√(π/2)，

即⎰(0, +∞) cos x /√x dx = ⎰(0, +∞) sin x /√x dx = √(π/2)．

8. 从⎰C √z ln z /(1 + z )2 dz 出发，其中C 是如图所示之周线，证明：

⎰(0, +∞) √x ln x /(1 + x )2 dx = π，⎰(0, +∞) √x /(1 + x )2 dx = π/2．

【解】在割去原点及正实轴的z 平面上，√z ，ln z 都

能分出单值解析分支，√z 取在正实轴的上岸取正值

的那个分支，ln z 取在正实轴的上岸取实数值的那个

分支．记f (z ) = √z ln z /(1 + z )2 dz ．f (z )的有限奇点只有- 1，且- 1是f (z )的2阶极点．

Res[√z ln z /(1 + z )2; - 1] = lim z → - 1 ((1 + z )2 · f (z ))’

= lim z → - 1 (√z ln z )’ = lim z → - 1 (((1/2) ln z + 1 )√z /z )

= ((1/2) ln (- 1) + 1 )√(- 1)/(- 1)

= - ((1/2) πi + 1 )i = (1/2) π - i ．

当r < 1 < R 时，⎰C √z ln z /(1 + z )2 dz

= ⎰C (r ) + ⎰C (R ) + ⎰L (1) + ⎰L (2) = 2πi Res[√z ln z /(1 + z )2; -1] = 2π + π2 i ．

⎰L (1) √z ln z /(1 + z )2 dz = ⎰(r , R ) √x ln x /(1 + x )2 dx

→ ⎰(0, +∞) √x ln x /(1 + x )2 dx (当r → 0+，R →+∞时)

⎰L (2) √z ln z /(1 + z )2 dz = ⎰(R , r ) (-√x )(ln x + 2πi )/(1 + x )2 dx

= ⎰(r , R ) (√x ln x )/(1 + x )2 dx + 2πi ⎰(r , R )√x /(1 + x )2 dx

→ ⎰(0, +∞) √x ln x /(1 + x )2 dx + 2πi ⎰(0, +∞) √x /(1 + x )2 dx (当r → 0+，R →+∞时)． 因为z · √z ln z /(1 + z )2 → 0 (当| z |→ +∞时)，

故⎰C(R) √z ln z/(1 + z)2dz→ 0 (当R → +∞时)．

因为z ·√z ln z/(1 + z)2 → 0 (当| z |→ 0时)，

故⎰C(r) √z ln z/(1 + z)2dz→ 0 (当r → 0时)．

所以，⎰L(1)+ ⎰L(2)→π/2 -i (当r→ 0+，R→+∞时)．

故2⎰(0, +∞)√x ln x/(1 + x)2dx + 2πi⎰(0, +∞) √x /(1 + x)2dx = 2π + π2i．

所以，⎰(0, +∞)√x ln x/(1 + x)2dx = π，⎰(0, +∞)√x /(1 + x)2dx = π/2．

9. 证明：I = ⎰(0, 1) 1/((1 + x2)(1 -x2)1/2) dx = π/23/2．

在割线的上岸(1 -z2)1/2取正值的那一支．

因i和-i都是f(z)的一阶极点，故

Res[ f(z); i] = 1/(2z (1 -z2)1/2)|z = i= -i/23/2．

Res[ f(z); i] = 1/(2z (1 -z2)1/2)|z = –i= -i/23/2．

若x在上岸，则f(x) = 1/((1 + x2)(1 -x2)1/2)；

若x在下岸，则f(x) = e-i π/((1 + x2)(1 -x2)1/2)；

⎰L(1) f(z) dz = ⎰[– 1 + r, 1 –r] f(x) dx．

⎰L(2) f(z) dz = ⎰[– 1 + r, 1 –r] f(x) dx．

因为lim z→–1 (1 + z) f(z) = 0，lim z→ 1 (1 -z) f(z) = 0，

故⎰S(r) f(z) dz→ 0，⎰T(r) f(z) dz→ 0 (as r → 0)．

因为lim z→∞z f(z) = 0，故⎰C(R) f(z) dz→ 0 (as R → +∞)．

故⎰L(1) f(z) dz + ⎰L(2) f(z) dz→ (2πi)(Res[ f(z); i] + Res[ f(z); -i]) (as r→ 0+，R→+∞)．所以2⎰(– 1, 1) f(x) dx = (2πi)(Res[ f(z); i] + Res[ f(z); -i]) = (2πi)(-i/23/2) = 2π/23/2．

故⎰(– 1, 1) f(x) dx = π/23/2．

10. 证明方程e z-λ= z ( λ> 1 )在单位圆| z | < 1内恰有一个根，且为实根．【解】在单位圆周C : | z | = 1上，设z = x + i y，则z-λ= (x -λ) + i y，

故| e z-λ| = | e (x -λ) + i y | = | e x -λ| < 1 = | z |，

由Rouché定理，N(z - e z-λ, C) = N(z, C) = 1．

故z - e z-λ = 0在单位圆内恰有一个根．

设f(x) = x - e x-λ，x∈ ．因f(- 1) = (- 1)- e-1 -λ < 0，f(1) = 1- e 1 -λ > 0，

故x - e x-λ = 0在区间(- 1, 1)内有根．

所以方程e z-λ= z ( λ> 1 )在单位圆| z | < 1内的唯一根为实根．

[原题是错题．例如c = 1/2，λ= 2，则∀z∈ ，当| z | < 1时，

| c z-λ| = | exp((z-λ) Ln c)| = | exp(( z– 2)(ln| 1/2| + 2kπi)) | = e (2 –z)ln2 > 1 > | z |．]

11. 证明方程e z- eλz n= 0 ( λ> 1 )在单位圆| z | < 1内有n个根．

【解】在单位圆周C : | z | = 1上，| e z| = e Re(z)≤ e | z |≤ e < eλ= | eλz n |，

由Rouché定理，N(eλz n- e z, C) = N(eλz n, C) = N(z n, C) = n．

12. 若f(z)在周线C内部除有一个一阶极点外解析，且连续到C，在C上| f(z) | = 1，证明f(z) = a ( | a | > 1 )在C内部恰好有一个根．

【解】考虑圆K = { z∈ | | z–a | < | a |}．