# 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

$F(s) = \int_0^\infty f(t) e^{-st}\,dt$

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

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

## Proof

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

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

thus:

$\lim_{s \to \infty} sF(s)=\lim_{s \to \infty} [f(0^-)+\int_{t=0^-}^{\infty}e^{-st}f^{'}(t)dt]$

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

$\lim_{s \to \infty} [\int_{t=0^-}^{\infty}e^{-st}f^{'}(t)dt] =\lim_{s \to \infty}\{\lim_{\epsilon \to 0^+}[\int_{t=0^-}^{\epsilon}e^{-st}f^{'}(t)dt] + \lim_{\epsilon \to 0^+}[\int_{t=\epsilon}^{\infty}e^{-st}f^{'}(t)dt]\}$ In the first expression where 0<t<0+, e−st=1. In the second expression, the order of integration and limit-taking can be changed. Also $\lim_{s \to \infty}e^{-st}(t)$ where 0+<t<∞ is zero. Therefore:[3]

\begin{align} \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{align} By substitution of this result in the main equation we get:

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