find a function $f(t)$ such that $\mathscr{L}\{f(t)\}(s)=f(s)/s$

Question

find a function $f(t)$ such that $\mathscr{L}\{f(t)\}(s)=f(s)/s$

Does such a function $f(t)$ exist?

This came up while I was playing around with p-adic Norms.

Answer 1

As shown in almost any book on the Laplace transform, we have that $$ \mathscr{L}\left[\int\limits_0^t f(r)\mathrm{d}r\right](s)=\frac{F(s)}{s} + \frac{1}{s} \int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r \iff \frac{F(s)}{s} = \mathscr{L}\left[\int\limits_0^t f(r)\mathrm{d}r\right](s) - \frac{1}{s} \int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r $$ and then using the fact that in this case you have $F(s)= f(s)$ we have $$ \begin{split} f(t)& =\mathscr{L}^{-1}\big[\mathscr{L}[f(t)](s)\big](t)\\ & = \mathscr{L}^{-1}\left[\mathscr{L}\left[\int\limits_0^t f(r)\mathrm{d}r\right](s) - \frac{1}{s} \int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r\right](t)\\ & = \int\limits_0^t f(r)\mathrm{d}r - H(t)\int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r \end{split} $$ where $H(t)$ is the Heaviside function. Thus you have an integral equation which can be reduced by differentiation to the following first order linear ordinary differential equation: $$ \frac{\mathrm{d}}{\mathrm{d} t}f(t)= f(t) - \delta (t)\int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r\label{1}\tag{1} $$ where $\delta(t)$ is the Dirac delta distribution.

Solution to equation \eqref{1}.

First note that \eqref{1} is linear since $$ c_\infty=\int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r\label{ic}\tag{IC}. $$ should be considered as a datum. Said that, we start seeking the solution of \eqref{1} in the form $f(t)= f_o(t) + f_p(t)$, where $f_p(t)$ is a particular solution found as described in this answer and $f_o(t)$ is the solution of the associated homogeneous equation $$ \frac{\mathrm{d}}{\mathrm{d} t}f_o(t)= f_o(t).\label{2}\tag{2} $$ We have $$ \begin{split} f_o(t) &=c e^t\\ f_p(t) &=c_\infty H(t)e^t \end{split} $$ where $c$ is a constant to be determined: and in order to determine it, we use the integral condition \eqref{ic}, obtaining $$ c_\infty = \int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r = \int\limits_{-\infty}^{0^+} f_o(r)\mathrm{d}r =c \int\limits_{-\infty}^{0^+} e^r\mathrm{d}r = c\big[e^t\big]_{-\infty}^{0+} = c $$ Finally, $$ f(t)= c_\infty e^t\big[1+H(t)\big] $$