※ 引述《nini851216 (解壓縮)》之銘言：
: 不知道從何下手
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這題的答案不對
f(x) = x^a / {[x - b]^2 + c^2}
∞
I = ∫f(x)dx
0
定義a + bi = rexp(iθ) 其中0 < α < 2π
(1)當c =/= 0
可知分母[x - b]^2 + c^2並無正實數根
[1 - exp(i2πa)]I
= 2πi[(r)^(a-1)exp(iθ(a-1)) - (r)^(a-1)exp(i(2π-θ)(a-1))]/(2ci)
=> I = [π/sin(πa)](1/c)r^(a-1)sin((θ-π)(a-1))
(2)當c = 0且b < 0，r = -b，θ = π
[1 - exp(i2πa)]I
= 2πi[(a-1)r^(a-2)exp(iπ(a-2))]
=> I = -[π/sin(πa)](a-1)r^(a-2)
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