title: Initial value theorem
text: # 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.
  
  Let
  
      $F(s)=\int _{0}^{\infty }f(t)e^{-st}\,dt$
  
  be the (one-sided) Laplace transform of ƒ(t). If $f$ is bounded on $(0,\infty )$ (or if just $f(t)=O(e^{ct})$) and $\lim _{t\to 0^{+}}f(t)$ exists then the initial value theorem says
  
      $\lim _{t\,\to \,0}f(t)=\lim _{s\to \infty }{sF(s)}.$
  
  ## Proofs
  
  ### Proof using dominated convergence theorem and assuming that function is bounded
  
  Suppose first that $f$ is bounded, i.e. $\lim _{t\to 0^{+}}f(t)=\alpha$. A change of variable in the integral $\int _{0}^{\infty }f(t)e^{-st}\,dt$ shows that
  
      $sF(s)=\int _{0}^{\infty }f\left({\frac {t}{s}}\right)e^{-t}\,dt$.
  
  Since $f$ is bounded, the Dominated Convergence Theorem implies that
  
      $\lim _{s\to \infty }sF(s)=\int _{0}^{\infty }\alpha e^{-t}\,dt=\alpha .$
  
  ### Proof using elementary calculus and assuming that function is bounded
  
  Of course we don't really need DCT here, one can give a very simple proof using only elementary calculus:
  
  Start by choosing $A$ so that $\int _{A}^{\infty }e^{-t}\,dt<\epsilon$, and then note that $\lim _{s\to \infty }f\left({\frac {t}{s}}\right)=\alpha$ uniformly for $t\in (0,A]$.
  
  ### Generalizing to non-bounded functions that have exponential order
  
  The theorem assuming just that $f(t)=O(e^{ct})$ follows from the theorem for bounded $f$:
  
  Define $g(t)=e^{-ct}f(t)$. Then $g$ is bounded, so we've shown that $g(0^{+})=\lim _{s\to \infty }sG(s)$. But $f(0^{+})=g(0^{+})$ and $G(s)=F(s+c)$, so
  
      $\lim _{s\to \infty }sF(s)=\lim _{s\to \infty }(s-c)F(s)=\lim _{s\to \infty }sF(s+c)=\lim _{s\to \infty }sG(s),$
  
  since $\lim _{s\to \infty }F(s)=0$.
  
  