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

(more info)

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

(Derivation)

Suppose that the quotient

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

is Laplace-transformable. It follows easily that also

is such. According to the parent entry, we may write

$\displaystyle \mathcal{L}^{-1}\left\{G'(s)\right\} \,=\, -t\,g(t) \,=\, -f(t) \,=\, \mathcal{L}^{-1}\left\{-F(s)\right\}.$

Therefore

$\displaystyle G'(s) \,=\, -F(s),$

whence

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

(1)

where $F^{(-1)}(s)$ means any antiderivative of

. Since each Laplace transformed function vanishes in the infinity $s = \infty$ and thus $G(\infty) = 0$ , the equation (1) implies

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

and therefore

$\displaystyle G(s) \,=\, F^{(-1)}(\infty)\!-\!F^{(-1)}(s) \,=\, \int_s^\infty F(u)\,du.$

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

$\displaystyle \mathcal{L}\left\{\frac{\sin{t}}{t}\right\} = \int_s^{\,\infty}\!... ...fty}\!\arctan{u} \,=\, \frac{\pi}{2}\!-\!\arctan{s} \,=\, {\mathrm{arccot}}{s}.$