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