In mathematical analysis, the initial value theorem is a theorem used to relate frequency domain expressions to the time domain behavior as time approaches zero.
It is also known under the abbreviation IVT.
Let
                    F        (        s        )        =                  ∫                      0                                ∞                          f        (        t        )                  e                      −            s            t                                  d        t                be the (one-sided) Laplace transform of ƒ(t). The initial value theorem then says
                              lim                      t            →            0                          f        (        t        )        =                  lim                      s            →            ∞                                    s          F          (          s          )                .                        Based on the definition of Laplace transform of derivative we have:
                    s        F        (        s        )        =        f        (                  0                      −                          )        +                  ∫                      t            =                          0                              −                                                          ∞                                    e                      −            s            t                                    f                                                  ′                                      (        t        )        d        t                thus:
                              lim                      s            →            ∞                          s        F        (        s        )        =                  lim                      s            →            ∞                          [        f        (                  0                      −                          )        +                  ∫                      t            =                          0                              −                                                          ∞                                    e                      −            s            t                                    f                                                  ′                                      (        t        )        d        t        ]                But                               lim                      s            →            ∞                                    e                      −            s            t                                   is indeterminate between t=0− to t=0+; to avoid this, the integration can be performed in two intervals:
                              lim                      s            →            ∞                          [                  ∫                      t            =                          0                              −                                                          ∞                                    e                      −            s            t                                    f                                                  ′                                      (        t        )        d        t        ]        =                  lim                      s            →            ∞                          {                  lim                      ϵ            →                          0                              +                                                    [                  ∫                      t            =                          0                              −                                                          ϵ                                    e                      −            s            t                                    f                                                  ′                                      (        t        )        d        t        ]        +                  lim                      ϵ            →                          0                              +                                                    [                  ∫                      t            =            ϵ                                ∞                                    e                      −            s            t                                    f                                                  ′                                      (        t        )        d        t        ]        }                In the first expression,
                              e                      −            s            t                          =        1                          w          h          e          n                                  0                      −                          <        t        <                  0                      +                          .                In the second expression, the order of integration and limit-taking can be changed. Also
                              lim                      s            →            ∞                                    e                      −            s            t                          (        t        )        =        0                          w          h          e          r          e                                  0                      +                          <        t        <        ∞        .                Therefore:
                                                                                          lim                                      s                    →                    ∞                                                  [                                  ∫                                      t                    =                                          0                                              −                                                                                                  ∞                                                                    e                                      −                    s                    t                                                                    f                                                                                  ′                                                                      (                t                )                d                t                ]                                                            =                                  lim                                      s                    →                    ∞                                                  {                                  lim                                      ϵ                    →                                          0                                              +                                                                                            [                                  ∫                                      t                    =                                          0                                              −                                                                                                  ϵ                                                                    f                                                                                  ′                                                                      (                t                )                d                t                ]                }                +                                  lim                                      ϵ                    →                                          0                                              +                                                                                            {                                  ∫                                      t                    =                    ϵ                                                        ∞                                                                    lim                                      s                    →                    ∞                                                  [                                  e                                      −                    s                    t                                                                    f                                                                                  ′                                                                      (                t                )                d                t                ]                }                                                                                                  =                f                (                t                )                                                      |                                                        t                    =                                          0                                              −                                                                                                  t                    =                                          0                                              +                                                                                            +                0                                                                                                  =                f                (                                  0                                      +                                                  )                −                f                (                                  0                                      −                                                  )                +                0                                                            By substitution of this result in the main equation we get:
                              lim                      s            →            ∞                          s        F        (        s        )        =        f        (                  0                      −                          )        +        f        (                  0                      +                          )        −        f        (                  0                      −                          )        =        f        (                  0                      +                          )