# 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

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

Therefore

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

whence

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

where $F^{(-1)}(s)$ means any antiderivative of $F(s)$.  Since each Laplace transformed function  vanishes in the infinity  $s=\infty$  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

 $\displaystyle\mathcal{L}\left\{\frac{f(t)}{t}\right\}\,=\,\int_{s}^{\infty}F(u% )\,du.$ (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

 $\mathcal{L}\left\{\frac{\sin{t}}{t}\right\}=\int_{s}^{\,\infty}\!\frac{1}{u^{2% }+1}\,du=\operatornamewithlimits{\Big{/}}_{\!\!\!s}^{\,\quad\infty}\!\arctan{u% }\,=\,\frac{\pi}{2}\!-\!\arctan{s}\,=\,\operatorname{arccot}{s}.$

Thus we have

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

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

Title Laplace transform of $\frac{f(t)}{t}$ LaplaceTransformOffracftt 2014-03-08 15:45:15 2014-03-08 15:45:15 pahio (2872) pahio (2872) 8 pahio (2872) Derivation msc 44A10 FundamentalTheoremOfCalculusClassicalVersion SubstitutionNotation CyclometricFunctions