When to use $e^{-iwt}$ and $e^{iwt}$ for Fourier transform?

Question

The Fourier tranform of message signal is defined as:

$$ M(w) = \mathcal{F}\{m(t)\} = \int^\infty _{-\infty} m(t).e^{-iwt}dt $$

and the reverse transform is defined as:

$$ m(t) = \mathcal{F}\{M(w)\} = \int^\infty _{-\infty} M(w).e^{iwt}dw $$

My question is that is that should $m(t)$ be necessarily multiplied with $e^{-iwt}$? What if I multiply it with $e^{iwt}$? Will the tranformation work? Why or why not?

Answer 1

Heuristically, $$ \int_{-\infty}^{\infty} e^{i\omega t} \left( \int_{-\infty}^{\infty} e^{-i\omega T} f(T) \, dT \right) \, d\omega = \int_{-\infty}^{\infty} f(T) \left( \int_{-\infty}^{\infty} e^{i\omega(t-T)} \, d\omega \right) \, dT \\ = 2\pi \int_{-\infty}^{\infty} \delta(t-T) f(T) \, dT = 2\pi f(t). $$ (The $2\pi$ is required; you can check it is necessary by computing the Fourier transform of $e^{-t^2/2}$.)

The (not-rigorous) idea here is that $e^{i\omega(t-T)}$ has a small integral unless $t=T$, when it has a very large integral.