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

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

Solutions Collecting From Web of "Fourier Transform Proof $ \mathcal F(f(x)g(x))=(\frac{1}{2\pi})F(s)*G(s)$"

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.