Suppose $$F(k) = \frac{1}{2\pi}\int f(x) e^{ikx} dx$$ and
$$G(k) = \frac{1}{2\pi}\int g(x)e^{ikx} dx$$
Where $F(K),G(K)$ are Fourier transforms. Then how can I write the convolution of $F$ and $G$ so that
$$F *G = \frac{1}{2\pi}\int f(x)g(x)e^{ikx} dx$$
Where $*$ is the convolution?
I know that if $$f(x) = \int F(K)e^{-ikx} dx$$ $$g(x) = \int G(K)e^{-ikx} dx$$
Then $$\frac{1}{2\pi}\int g(x-x')f(x')dx' = f(x) = \int F(K)G(K)e^{-ikx} dx$$
What would the expression of the Fourier transform of $f(x)g(x)$ be in terms of $F(K)$ and $G(K)$