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Problem of the Week #42 - January 14th, 2013

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Chris L T521

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Jan 26, 2012
995
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:
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.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by Sudharaka. You can find his solution below.

\[F(s)G(s)=\int_0^{\infty}\int_0^{\infty}e^{-s(x+y)}f(x)g(y)\,dx\,dy\]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,
\[F(s)G(s)=\int_0^{\infty}e^{-st}h(t)\,dt\]
 
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