14.4: The Second Integration Theorem (Dividing a Function by t)

This theorem looks very like the first integration theorem, but "the other way round". It is

\[\textbf{L} \left(\frac{y(t)}{t} \right) = \int_s^\infty \bar{y}(x)dx.\]

I'll leave it for the reader to derive the theorem. Here I just give an example of its use. Whereas the first integration theorem is most useful in finding inverse transforms, the second integration theorem is more useful for finding direct transforms.

Example \(\PageIndex{1}\)

 Calculate

\[\textbf{L}\left(\frac{\sin at}{t} \right).\]

Solution

This means calculate

\[\int_0^\infty \frac{e^{-st}\sin at}{t}dt.\]

While this integral can no doubt be done, you may find it a bit daunting, and the second integration theorem provides an alternative way of doing it, resulting in an easier integral.

Note that the right hand side of equation 14.4.1 is a function of \(s\), not of \(x\), which is just a dummy variable. The function \(\bar{y}(x)\) is the Laplace transform, with \(x\) as argument, of \(y(t)\). In our particular case, \(y(t)\) is \(\sin at\), so that, from the table, \(\bar{y}(x)=\frac{a}{a^2+x^2}\). The second integration theorem, then, tells us that \(\textbf{L} \left( \frac{\sin at}{t}\right)=\int_s^\infty \frac{a}{a^2+x^2}dx\). This is a much easier integral. It is \(\left[\tan^{-1} \left(\frac{x}{a} \right) \right]_s^\infty = \frac{\pi}{2} - \tan^{-1} \left( \frac{s}{a} \right) = \tan^{-1} \left(\frac{a}{s}\right)\). You may want to add this result to your table of Laplace integrals. Indeed, you may already want to expand the table considerably by applying both integration theorems to several functions.

Contributor

• Jeremy Tatum (University of Victoria, Canada)