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I need to prove this:

$$ \mathcal F(f(x)g(x))=(\frac{1}{2\pi})F(s)*G(s)$$

So far, I believe I have to use the Fourier transform standard equation

$$ \mathcal F(f(x))=\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-isx}\,dx $$

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You can use the inverse Fourier transform $\mathcal F^{-1}$ to simplify the problem first

$$f(x)g(x)=\frac{1}{2\pi}\mathcal F^{-1}\left(F(s)*G(s)\right)$$ where

$$f(x)=\mathcal F^{-1}(F(s))=\int_{-\infty}^{\infty}ds F(s)e^{isx}$$

now, using the definition of convolution

$$F(s)*G(s)=\int_{-\infty}^{\infty} ds’F(s’)G(s-s’)$$

Applying the inverse Fourier transform

$$\mathcal F^{-1}(F(s)*G(s))=\int_{-\infty}^{\infty}ds\int_{-\infty}^{\infty} ds’F(s’)G(s-s’)e^{isx}$$

multiplying and dividing with $e^{is’x}$

$$\int_{-\infty}^{\infty}ds\int_{-\infty}^{\infty} ds’F(s’)G(s-s’)e^{i(s-s’)x}e^{isx}=f(x)g(x)$$

Which is what you wanted to demonstrate, up to the $\frac{1}{2\pi}$ prefactor in your question.

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