MHB Can You Master This Week's Challenge on Laplace Transforms and Convolution?

[Chris L T521]

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

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Problem: Let $f,g:[0,\infty)\rightarrow\mathbb{R}$ be two functions, and let $F(s)$ and $G(s)$ denote their Laplace Transforms. Show that
\[F(s)G(s)=\int_0^{\infty} e^{-st}h(t)\,dt\]
where $h(t) = \int_0^t f(t-\tau)g(\tau)\,d\tau$ (the convolution of $f$ with $g$).

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Hint:

Spoiler

Start with the double integral
\[F(s)G(s)=\int_0^{\infty}\int_0^{\infty}e^{-s(x+y)}f(x)g(y)\,dx\,dy\]
Make the change of variables $t=x+y$, $y=\tau$ and then change the order of integration.

 

[Chris L T521]

This week's problem was correctly answered by Sudharaka. You can find his solution below.

Spoiler

\[F(s)G(s)=\int_0^{\infty}\int_0^{\infty}e^{-s(x+y)}f(x)g(y)\,dx\,dy\]Substitute \(t=x+y\) and \(y=\tau\) and we get,
\[F(s)G(s)=\int_0^{\infty}\int_\tau^{\infty}e^{-st}f(t-\tau)g(\tau)\,dt\,d\tau\]
By changing the order of integration we get,
\[F(s)G(s)=\int_0^{\infty}\int_0^t e^{-st}f(t-\tau)g(\tau)\,d\tau\,dt=\int_0^{\infty}e^{-st}\left[\int_0^t f(t-\tau)g(\tau)\,d\tau\right]\,dt\]
Taking \(h(t)=\int_0^t e^{-st}f(t-\tau)g(\tau)\,d\tau\) we get,
\[F(s)G(s)=\int_0^{\infty}e^{-st}h(t)\,dt\]