Laplace transform of $\frac{f(t)}{t}$

Suppose that the quotient

$$\frac{f(t)}{t}:=g(t)$$

is Laplace-transformable (http://planetmath.org/LaplaceTransform).  It follows easily that also $f(t)$ is such.  According to the parent entry (http://planetmath.org/LaplaceTransformOfTnft), we may write

$${ℒ}^{-1}\left\{G^{\prime }(s)\right\}=-tg(t)=-f(t)={ℒ}^{-1}\left\{-F(s)\right\}.$$

Therefore

$$G^{\prime }(s)=-F(s),$$

whence

$G(s)=-F^{(-1)}(s)+C$(1)

where $F^{(-1)}(s)$ means any antiderivative of $F(s)$.  Since each Laplace transformed function vanishes in the infinity  $s=\mathrm{\infty }$  and thus  $G(\mathrm{\infty })=0$,  the equation (1) implies

$$C=F^{(-1)}(\mathrm{\infty })$$

and therefore

$$G(s)=F^{(-1)}(\mathrm{\infty })-F^{(-1)}(s)={\int }_s^{\mathrm{\infty }}F(u)𝑑u.$$

We have obtained the result

$ℒ\left\{{\displaystyle \frac{f(t)}{t}}\right\}={\displaystyle {\int }_s^{\mathrm{\infty }}}F(u)𝑑u.$(2)

Application.  By the table of Laplace transforms, $ℒ\left\{\sin t\right\}={\displaystyle \frac{1}{s^2+1}}.$  Accordingly the formula (2) yields

$$ℒ\left\{\frac{\sin t}{t}\right\}={\int }_s^{\mathrm{\infty }}\frac{1}{u^2+1}𝑑u=\underset{s}{\overset{\mathrm{\infty }}{/}}\arctan u=\frac{\pi }{2}-\arctan s=\mathrm{arccot}s.$$

Thus we have

$ℒ\left\{{\displaystyle \frac{\sin t}{t}}\right\}=\mathrm{arccot}s=\arctan {\displaystyle \frac{1}{s}}.$(3)

This result is derived in the entry Laplace transform of sine integral in two other ways.