求定积分(分步积分法),

1个回答

  • ∫﹙1→∞﹚td[e^﹙-st﹚]

    =t[e^﹙-st﹚]|﹙1→∞﹚- ∫﹙1→∞﹚[e^﹙-st﹚]dt

    =【0-e^﹙-s﹚】-﹙-1/s﹚ ∫﹙1→∞﹚[e^﹙-st﹚]d﹙-st﹚

    =-e^﹙-s﹚+1/s×e^﹙-st﹚|﹙1→∞﹚

    =-e^﹙-s﹚+1/s×[0-e^﹙-s﹚]

    =-e^﹙-s﹚﹙1+1/s﹚