广义积分习题

广义积分习题1.计算dxxx2211；2.计算dxxx212；3.计算10lnxdx；4.计算4342cosxdx；5.判定dxxxn12ln（n0）的敛散性。6.判定dxxx1021ln的敛散性。解答1.计算dxxx2211；dtttttdxxx2322tansectansec11（x=sect）632.2.计算dxxx212；dxxxdxxxdxxx02022121212bbbbxxdxdxxx0220202|)1ln(lim111lim12，发散故广义积分dxxx212发散。3.计算10lnxdx；]1|ln[limlnlimln1101010dxxxxxxdxxdx1)]1(ln[lim04.计算4342cosxdx；43222424342coscoscosxdxxdxxdxttttxxdxxdx42422242|)(tanlimseclimcos)1(tanlim2tt发散故广义积分4342cosxdx发散。5.判定dxxxn12ln（n0）的敛散性。21121223211lnlimlnlimlnlimxxxnxxxxxnxnxnx0lnlim2211xxnnx所以dxxxn12ln收敛6.判定dxxx1021ln的敛散性两个瑕点x=0,x=1dxxxdxxxdxxx121221021021ln1ln1ln(1)dxxx21021ln瑕点x=00lnlim1lnlim2102210xxxxxxx故dxxx21021ln收敛（2）dxxx12121ln瑕点x=10)1(lnlim21)1)(1(lnlim1ln)1(lim2112112211xxxxxxxxxxx故dxxx12121ln收敛因此广义积分dxxx1021ln收敛。