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14.3: The First Integration Theorem

  • Page ID
    5503
  • [ "article:topic", "First Integration Theorem", "inverse Laplace transform", "authorname:tatumj" ]

    First Integration Theorem

    The theorem is:

    \[\textbf{L} \int_0^t y(x)dx = \frac{\bar{y}(s)}{s}.\]

    Before deriving this theorem, here's a quick example to show what it means. The theorem is most useful, as in this example, for finding an inverse Laplace transform, i.e.

    \[\textbf{L}^{-1} \frac{\bar{y} (s)}{s} = \int_0^t y(x)dx.\]

    Example \(\PageIndex{1}\)

    Calculate

    \[\textbf{L}^{-1} \frac{1}{s(s-a)}.\]

    Solution

    From the table, we see that \(\textbf{L}^{-1} \frac{1}{s-a}=e^{at}\). The integration theorem tells us that

    \[\textbf{L}^{-1} \frac{1}{s(s-a)}=\int_o^t e^{ax}dx = (e^{at}-1)/a.\]

    You should now verify that this is the correct answer by substituting this in Equation 14.1.2 and integrating – or (and!) using the table of Laplace transforms. 

    Proof

    The proof of the theorem is just a matter of integrating by parts. Thus

    \[\begin{align} \textbf{L}\int_0^t y(x)dx & = \int_0^\infty \left( \int_0^t y(x)dx \right) e^{-st}dt = -\frac{1}{s}\int_0^\infty \left( \int_0^t y(x)dx \right) d\left( e^{-st} \right) \\ &= \left[ -\frac{1}{s}e^{-st} \int_0^t y(x)dx \right]^\infty_{t=0} + \frac{1}{s} \int_0^\infty e^{-st}y(t)dt. \end{align}\]

    The expression in brackets is zero at both limits, and therefore the theorem is proved.

    Contributor