解:∫(0,1)lnx/(1-x)dx= 令y=1-x,则dx=-dy
∫(1,0)ln(1-y)/y*(-dy)=
∫(0,1)ln(1-y)/y*dy
考虑用级数法,由于
ln(1+u)=u-u²/2+u³/3-……
故
ln(1-y)=-y-y²/2-y³/3-……
ln(1-y)/y=-1-y/2-y²/3-y³/4-……
于是
原积分=∫(0,1)ln(1-y)/y*dy=
-∫(0,1)(1+y/2+y²/3+y³/4+……)dy=
-(1/1²+1/2²+1/3²+1/4²+……)
=-π²/6