|
Suppose that the quotient $$\frac{f(t)}{t} \,:=\; g(t)$$ is Laplace-transformable. It follows easily that also $f(t)$ is such. According to the parent entry, we may write $$\mathcal{L}^{-1}\left\{G'(s)\right\} \,=\, -t\,g(t) \,=\, -f(t) \,=\, \mathcal{L}^{-1}\left\{-F(s)\right\}.$$ Therefore $$G'(s) \,=\, -F(s),$$ whence
 |
(1) |
where $F^{(-1)}(s)$ means any antiderivative of $F(s)$ . Since each Laplace transformed function vanishes in the infinity $s = \infty$ and thus $G(\infty) = 0$ , the equation (1) implies $$C \,=\, F^{(-1)}(\infty)$$ and therefore $$G(s) \,=\,
F^{(-1)}(\infty)\!-\!F^{(-1)}(s) \,=\, \int_s^\infty F(u)\,du.$$ We have obtained the result
 |
(2) |
Application. By the table of Laplace transforms, $\displaystyle\mathcal{L}\left\{\sin{t}\right\} = \frac{1}{s^2+1}.$ Accordingly the formula (2) yields $$\mathcal{L}\left\{\frac{\sin{t}}{t}\right\} = \int_s^{\,\infty}\!\frac{1}{u^2+1}\,du = \sijoitus{s}{\quad\infty}\!\arctan{u} \,=\, \frac{\pi}{2}\!-\!\arctan{s} \,=\, \arccot{s}.$$ Thus we have
 |
(3) |
This result is derived in the entry Laplace transform of sine integral in two other ways.
|
![[parent]](http://images.planetmath.org:8080/images/uparrow.png)