Calculate Laplace transformation of the product

Question

Laplace transformation of the derivative of the product equals?

$$L\left(\frac{d(f_1(y)f_2(y))}{dy}\right) = sf_1(s)f_2(s) - f_1(0)f_2(0) ?$$

I have meant not the convolution of the functions under the derivative operator, rather the derivative of their product. Thank you.

Answer 1

Let $g(y)=f_1(y) f_2(y)$, then we have

$$ \mathcal{L}(g') = sG(s)-g(0), $$

where

$$ G(s)=\mathcal{L}\left\{ f_1 f_2\right\}(s). $$

Added On the other hand, the Laplace of the convolution is

$$ \mathcal{L}\left\{ f_1*f_2 \right\}=\mathcal{L} \left\{ f_1\right\} \mathcal{L} \left\{ f_2\right\} .$$

There are not the same.

Answer 2

Let $ \mathcal{L}\{f(t)\}(s)=F(s)$ and $ \mathcal{L}\{g(t)\}(s)=G(s)$

So, we have:

$ \mathcal{L}\left\{ f(t)g(t) \right\}(s)= \frac{1}{2i\pi} \lim_{T \to \infty} \int\limits_{Re(w)-iT}^{Re(w)+iT} F(w)G(s-w)\,\mathrm dw, $

Or: you can take the Laplace Transform as a particular case of Fourier Transform.

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For more information see http://en.wikipedia.org/wiki/Laplace_transform