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

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#### Chris L T521

##### Well-known member
Staff member
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:
$F(s)G(s)=\int_0^{\infty}\int_0^{\infty}e^{-s(x+y)}f(x)g(y)\,dx\,dy$
Make the change of variables $t=x+y$, $y=\tau$ and then change the order of integration.

#### Chris L T521

##### Well-known member
Staff member
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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