When to use $e^{-iwt}$ and $e^{iwt}$ for Fourier transform? 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?  
 A: There are (too) many conventions for the definition of the Fourier transform. Some authors use a plus sign in the exponent; others use a minus sign. But whichever sign is used in the Fourier transform, the other sign must be used in the inverse Fourier transform; otherwise it won't be an inverse.
Also, to get an inverse, i.e., to ensure that applying the Fourier transform followed by the inverse transform gets you back to the original function (if that function was nicely behaved) a factor $2\pi$ is needed somewhere. Some authors simply divide by $2\pi$ the integral that defines the inverse transform; that gets rid of the $2\pi$ at the end of the calculation in Chappers's answer. Other authors prefer to divide by $\sqrt{2\pi}$ the integrals in the definitions of both the Fourier transform and the inverse transform. Still other authors choose instead to put a factor $2\pi$ into the exponents inside the integrals. 
Unfortunately, as far as I know, none of these conventions has established itself as clearly preferred over the others. So you just have to check the convention in each particular book or paper that you study. But (returning to your actual question) the signs of the exponents in the integrals will always be opposite for the Fourier transform and its inverse.
A: 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.
