设f(x,y)为连续函数,则积分∫下限1/n 上线n (1)
令t=1+1/x² 则关于t的积分上限是1+1/n² 积分下限是1+n²x=(t-1)^(-1/2) dx=-1/2(t-1)^(-3/2) dt
1-1/x²=2-t
原极限=∫(上限,下限)[-1/2(2-t)(t-1)^(-3/2)f(t)]dt
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