14.3: The First Integration Theorem

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.