广义微积分,dx/(lnx)^3从1到2的积分
答案:1 悬赏:20
解决时间 2021-01-19 08:34
- 提问者网友:心牵心
- 2021-01-19 03:56
广义微积分,dx/(lnx)^3从1到2的积分
最佳答案
- 二级知识专家网友:荒野風
- 2021-01-19 05:09
令t=lnx x=e^t dx=e^tdt
原式=∫(0,ln2) e^t/t^3dt
=(-1/2)*∫(0,ln2) e^td(1/t^2)
=(-1/2)*[e^t/t^2|(0,ln2)-∫(0,ln2) e^t/t^2dt]
=(-1/2)*[∫(0,ln2) e^td(1/t)+e^t/t^2|(0,ln2)]
=(-1/2)*[e^t/t|(0,ln2)-∫(0,ln2) e^t/tdt+e^t/t^2|(0,ln2)]
=(-1/2)*[e^t(1/t+1/t^2)|(0,ln2)-∫(0,ln2) e^td(lnt)]
=(1/2)*[(1/2)*e^tlnt|(0,ln2)+e^t(1/t+1/t^2)|(0,ln2)]
=(1/2)*[e^t*(lnt/2+1/t+1/t^2)|(0,ln2)]
=ln√(ln2)+1/ln2+1/(ln2)^2-lim(t->0+)[e^t*(t^2*lnt+2t+2)/4t^2]
=-∞
原式=∫(0,ln2) e^t/t^3dt
=(-1/2)*∫(0,ln2) e^td(1/t^2)
=(-1/2)*[e^t/t^2|(0,ln2)-∫(0,ln2) e^t/t^2dt]
=(-1/2)*[∫(0,ln2) e^td(1/t)+e^t/t^2|(0,ln2)]
=(-1/2)*[e^t/t|(0,ln2)-∫(0,ln2) e^t/tdt+e^t/t^2|(0,ln2)]
=(-1/2)*[e^t(1/t+1/t^2)|(0,ln2)-∫(0,ln2) e^td(lnt)]
=(1/2)*[(1/2)*e^tlnt|(0,ln2)+e^t(1/t+1/t^2)|(0,ln2)]
=(1/2)*[e^t*(lnt/2+1/t+1/t^2)|(0,ln2)]
=ln√(ln2)+1/ln2+1/(ln2)^2-lim(t->0+)[e^t*(t^2*lnt+2t+2)/4t^2]
=-∞
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