过程不算很详细,
(1) e^(iz)在原点的幂级数展开为1+iz+(iz)²/2+(iz)³/6+...
因此f(z) = (1+iz-e^(iz))/z² = 1/2+iz/6+...
可知a = lim{z → 0} f(z) = 1/2.
(2) 在定义f(0) = a以后,f(z)在整个复平面上解析.
由Cauchy积分定理,f(z)沿闭曲线D_A的积分∫{D_A} f(z)dz = 0.
(3) 曲线C_A可参数化为z = Ae^(it),t由0到π.
故∫{C_A} (1+iz)/z² dz = ∫{0,π} (1+iAe^(it))/(Ae^(it))² d(Ae^(it))
= ∫{0,π} (1+iAe^(it))/(Ae^(it))²·Aie^(it) dt
= i/A·∫{0,π} e^(-it)dt + ∫{0,π} (-1)dt
= i/A·(ie^(-iπ)-ie^(-i0))-π
= 2/A-π.
于是lim{A → +∞} ∫{C_A} (1+iz)/z² dz = -π.
(4) |∫{C_A} e^(iz)/z² dz| = |∫{0,π} e^(iAe^(it))/(Ae^(it))² d(Ae^(it))|
= |∫{0,π} e^(iAe^(it))/(Ae^(it)) dt|
≤ ∫{0,π} |e^(iAe^(it))/(Ae^(it))| dt
= ∫{0,π} |e^(-Asin(t)+iAcos(t))|/|Ae^(it)| dt (当b为实数,|e^(ib)| = 1).
= ∫{0,π} e^(-Asin(t))/A dt
≤ ∫{0,π} 1/A dt (Asin(t) ≥ 0,故e^(-Asin(t)) ≤ 1)
= π/A.
当A → +∞时π/A → 0,可得lim{A → +∞} |∫{C_A} e^(iz)/z² dz| = 0.
故lim{A → +∞} ∫{C_A} e^(iz)/z² dz = 0.
(5) 由(3)(4)的结果,可得lim{A → +∞} ∫{C_A} f(z)dz = -π.
又由0 = ∫{D_A} f(z)dz = ∫{C_A} f(z)dz+∫{-A,A} f(z)dz,
可得lim{A → +∞} ∫{-A,A} f(z)dz = π.
注意到∫{-A,A} f(z)dz = ∫{-A,0} f(z)dz+∫{0,A} f(z)dz
= ∫{0,A} f(-z)dz + ∫{0,A} f(z)dz
= ∫{0,A} f(z)+f(-z) dz
= ∫{0,A} (1+iz-e^(iz)+1-iz-e^(-iz))/z² dz
= 2∫{0,A} (1-cos(z))/z² dz.
因此∫{0,+∞} (1-cos(x))/x² dx = lim{A → +∞} ∫{0,A} (1-cos(z))/z² dz = π/2.
