第一题瑕积分怎么做

Question

如题所述

Answer 1

先求不定积分: ∫ 1/(x(1+ln(x)²)) dx = ∫ 1/(1+ln(x)²) d(ln(x)) = arctan(ln(x))+C.
于是瑕积分∫{0,1} 1/(x(1+ln(x)²)) dx
= lim{δ → 0+} ∫{δ,1} 1/(x(1+ln(x)²)) dx
= lim{δ → 0+} (arctan(ln(1))-arctan(ln(δ)))
= -lim{δ → 0+} arctan(ln(δ))
= π/2.