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Laplace transform of convolution

Posted December 29th, 2007 by Structure

Convolution of originals.
Let f,g be original function.
Then

$$f*g(t)=\int_{0}^{+\infty}f(\tau)g(t-\tau)d\tau=\int_{0}^{t}f(\tau)g(t-\tau)d\tau$$

is an original and

$$\mathcal{L}(f*g(t))(s)=\int_{0}^{+\infty}f*g(t)e^{-st}dt=\int_{0}^{+\infty}\int_{0}^{t}f(\tau)g(t-\tau)d\tau e^{-st}dt=$$
$$=\int_{0}^{+\infty}f(\tau)e^{-s\tau}(\int_{\tau}^{+\infty}g(t-\tau) e^{-s(t-\tau)}dt)d\tau=\int_{0}^{+\infty}f(\tau)e^{-s\tau}d\tau\int_{\tau}^{+\infty}g(t-\tau)e^{-s(t-\tau)}dt=$$
$$=\int_{0}^{+\infty}f(\tau)e^{-s\tau}d\tau\int_{0}^{+\infty}g(t')e^{-st'}dt'=F(s)G(s)$$
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