证明反常积分:∫b a dx/(x-a)^q 当0

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q=1时,原式=ln(x-a)[b~a]
=ln(b-a) - lim[x→a+] ln(x-a)
x→a+ ,x - a →0+ ,ln(x-a)→ - ∞
∴ln(b-a) - lim[x→a+] ln(x-a) = +∞
所以发散q≠1时原式 = (x-a)^(1-q) / (1-q) | [a,b]
= 1/(1-q) * { (b-a)^(1-q) - lim[x→a+] (x-a)^(1-q) }
q>1时x-a→0+,1-q