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[tex] \mathscr{F}\{f(t) g(t)\}=\mathcal{F}\{f(t)g(t)\}=\mathcal{F}\{f(t)\}\ast \mathcal{F}\{g(t)\}-\mathscr{F}\{f(t)\} \ast \mathscr{F}\{g(t)\} [/tex]

[tex] \mathscr{F}\{f(t) \ast g(t)\}=\mathcal{F}\{f(t)\ast g(t)\}=\mathcal{F}\{f(t)\}\mathcal{F}\{g(t)\}= \mathscr{F}\{f(t)\}\mathscr{F}\{g(t)\}[/tex]

and the same for the Laplace Transform.

However, my sources use the definition of the Fourier Transform as

[tex] \mathscr{F}\{f(t)\}=\mathcal{F}\{f(t)\}=\int_{-\infty}^{\infty}{e}^{-2i\pi\xi t}f(t) \mbox{d}t [/tex]

But my question is that if I were to use the definition,

[tex] \mathscr{F}\{f(t)\} = \mathcal{F}\{f(t)\}=\frac{1}{ \sqrt{2\pi} }\int_{-\infty}^{\infty}{e}^{-i\omega t}f(t) \mbox{d}t [/tex]

then will the convolution theorem for Fourier transforms still hold? There is one article that states that it does hold. However, when I checked Wikipedia, they say that the convolution theorem for Fourier transforms does not hold for the definition which uses angular frequency instead of ordinary frequency.

So, does the Convolution theorem for Fourier Transforms still hold for the definition involving omega instead of xi.