MHB How to show that $G(s) = \dfrac{F(s)}{s}$ for Laplace transforms of integrals?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: Suppose that $\displaystyle g(t) = \int_0^t f(\tau)\,d\tau$. If $G(s)$ and $F(s)$ are the Laplace transforms of $g(t)$ and $f(t)$ respectively, show that $G(s) = \dfrac{F(s)}{s}$.

Recall that $\displaystyle F(s) = \mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st}f(t)\,dt$.

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[Chris L T521]

This week's problem was correctly answered by BAdhi and Sudharaka.

Sudharaka's solution:

Spoiler

\[g(t) = \int_0^t f(\tau)\,d\tau\]

\begin{eqnarray}G(s)&=&\mathcal{L}\{g(t)\}\\&=&\int_0^{\infty} e^{-st}\int_0^t f(\tau)\,d\tau\,dt\\&=&\int_0^{\infty}\int_{0}^{t}e^{-st}f(\tau)\,d\tau\,dt\end{eqnarray}Note that, \(0<\tau<t<\infty\). Changing the order of integration we get,\begin{eqnarray}G(s)&=&\int_0^{\infty}\int_{\tau}^{\infty}e^{-st}f(\tau)\,dt\,d\tau\\&=&\int_0^{\infty}f(\tau)\int_{\tau}^{\infty}e^{-st}\,dt\,d\tau\\&=&\int_0^{\infty}f(\tau)\frac{e^{-s\tau}}{s}\,d\tau\\&=&\frac{1}{s}\int_0^{\infty}e^{-s\tau}f(\tau)\,d\tau\\\therefore G(s)=\frac{F(s)}{s}\end{eqnarray}

BAdhi's solution:

Spoiler

Let's consider the laplace transform of $g(t)$

$$G(s)=\mathcal{L}\{g(t)\}=\int_0^\infty e^{-st}g(t)\, dt$$since, $g(t)=\displaystyle \int_0^t f(\tau )\, d\tau$$$\begin{align*}
\mathcal{L}\{g(t)\}&=\int \limits_0^\infty e^{-st}g(t)\, dt \\ &=\int \limits_0^\infty e^{-st}\left[ \int_0^t f(\tau )\, d\tau \right] dt \\ &=\int \limits_{0}^{\infty}\int \limits_{0}^{t} e^{-st}f(\tau )\, d\tau dt\\ \end{align*}$$By switching variables, the boundaries of $t$ and $\tau$ changes as,$0<\tau<\infty$ and $\,\tau<t<\infty$then,$$
\mathcal{L}\{g(t)\}=\int \limits_0^\infty \int \limits_\tau^\infty e^{-st}f(\tau )\, dtd\tau $$since $f(\tau )$ is independent on $t$,$$\begin{align*}
\mathcal{L}\{g(t)\}&=\int \limits_{0}^{\infty}f(\tau )\left[ \int \limits_\tau^\infty e^{-st}\,dt \right] d\tau \\ &=\int \limits_0^\infty f(\tau ) \left[ \frac{e^{-st}}{-s}\right] _\tau^\infty \,d\tau \\ &=\int \limits_0^\infty f(\tau ) \left[ \frac{0-e^{-s\tau }}{-s}\right] \,d\tau (when s>0) \\&=\frac{\displaystyle \int \limits_0^\infty f(\tau ) e^{-s\tau }\,d\tau }{s}\\ &=\frac{\displaystyle \int \limits_0^\infty f(t) e^{-st }\,dt }{s}\\ &=\frac{F(s)}{s} \end{align*}$$$$\therefore \; G(s)=\frac{F(s)}{s}$$ for $s>0$