计算广义积分∫[0,1]ln[1/(1-x²)]dx
2个回答
先求不定积分
∫ ln(1/(1-x²))dx
=-∫ ln(1-x²)dx
=-xln(1-x²)-2∫ x²/(1-x²)dx
=-xln(1-x²)+2∫ (1-x²-1)/(1-x²)dx
=-xln(1-x²)+2∫ 1dx-2∫ 1/(1-x²)dx
=-xln(1-x²)+2x+ln(1-x)-ln(1+x)
=2x-ln(1+x)-ln[(1-x²)^x/(1-x)]
=2x-ln(1+x)-ln[(1+x)(1-x²)^x/(1-x²)]
=2x-ln(1+x)-ln[(1+x)(1-x²)^(x-1)] (1)
上式将0代入很简单,结果为0,关键是求x-->1-时的极限
下面计算：lim [x-->1-](1-x²)^(x-1)
=lim [x-->1-]e^[(x-1)ln(1-x²)]
=e^( lim [x-->1-] ln(1-x²)/(x-1)^-1 )
洛必达
=e^( lim [x-->1-] (-2x/(1-x²)) / -(x-1)^-2 )
=e^( lim [x-->1-] (2x(x-1)^2/(1-x²)))
=e^0=1
因此可得（1）当x-->1- 时的极限为：2-ln2-ln2=2-2ln2
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