# Initial value theorem

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.[1]

It is also known under the abbreviation IVT.

Let

${\displaystyle F(s)=\int _{0}^{\infty }f(t)e^{-st}\,dt}$

be the (one-sided) Laplace transform of ƒ(t). The initial value theorem then says[2]

${\displaystyle \lim _{t\to 0}f(t)=\lim _{s\to \infty }{sF(s)}.\,}$

## Proof

Based on the definition of Laplace transform of derivative we have:

${\displaystyle sF(s)=f(0^{-})+\int _{t=0^{-}}^{\infty }e^{-st}f^{'}(t)dt}$

thus:

${\displaystyle \lim _{s\to \infty }sF(s)=\lim _{s\to \infty }[f(0^{-})+\int _{t=0^{-}}^{\infty }e^{-st}f^{'}(t)dt]}$

But ${\displaystyle \lim _{s\to \infty }e^{-st}}$ is indeterminate between t=0 to t=0+; to avoid this, the integration can be performed in two intervals:

${\displaystyle \lim _{s\to \infty }[\int _{t=0^{-}}^{\infty }e^{-st}f^{'}(t)dt]=\lim _{s\to \infty }\{\lim _{\epsilon \to 0^{+}}[\int _{t=0^{-}}^{\epsilon }e^{-st}f^{'}(t)dt]+\lim _{\epsilon \to 0^{+}}[\int _{t=\epsilon }^{\infty }e^{-st}f^{'}(t)dt]\}}$

In the first expression,

${\displaystyle e^{-st}=1\qquad \mathrm {when} \qquad 0^{-}

In the second expression, the order of integration and limit-taking can be changed. Also

${\displaystyle \lim _{s\to \infty }e^{-st}(t)=0\qquad \mathrm {where} \qquad 0^{+}

Therefore:[3]

{\displaystyle {\begin{aligned}\lim _{s\to \infty }[\int _{t=0^{-}}^{\infty }e^{-st}f^{'}(t)dt]&=\lim _{s\to \infty }\{\lim _{\epsilon \to 0^{+}}[\int _{t=0^{-}}^{\epsilon }f^{'}(t)dt]\}+\lim _{\epsilon \to 0^{+}}\{\int _{t=\epsilon }^{\infty }\lim _{s\to \infty }[e^{-st}f^{'}(t)dt]\}\\&=f(t)|_{t=0^{-}}^{t=0^{+}}+0\\&=f(0^{+})-f(0^{-})+0\\\end{aligned}}}

By substitution of this result in the main equation we get:

${\displaystyle \lim _{s\to \infty }sF(s)=f(0^{-})+f(0^{+})-f(0^{-})=f(0^{+})}$