# Positive or negative sign on Fourier transform formula [duplicate]

I have seen both the formula of Fourier transform with positive and negative sign on exponential as $$X(\omega)=\int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt$$ and $$X(\omega)=\int_{-\infty}^{\infty} x(t)e^{j\omega t}dt$$ I am confused which one is the correct formula. I also solved for Fourier transform by taking the following example $$x(t)=\begin{cases} 1, \hspace{5mm} \text{for} \hspace{2mm} |t|<1 \\0, \hspace{5mm} \text{for} \hspace{2mm} |t|>1 \end{cases}$$ and got the same result as $$X(\omega)=\begin{cases} 2\frac{\text{sin}\omega}{\omega}, \hspace{5mm} \text{when} \hspace{2mm} \omega \neq 0 \\2, \hspace{13mm} \text{when} \hspace{2mm} \omega = 0\end{cases}$$ Can anyone explain whether both the formula for Fourier transform are correct or not?

The definition with the negative in the exponent is the accepted definition of the Fourier transform... however, this is an arbitrary choice. It could just as easily be defined with $e^{jw}$ and the inverse transform with $e^{-jw}$.
• in fact, while they are negatives of each other, there is no other difference between $-j$ and $+j$. both have equal claim to being the $\sqrt{-1}$. – robert bristow-johnson Oct 5 '15 at 0:06
• what will happen if we consider only the positive value of $t$? i.e. to say range of integration is consider from $0$ to $\infty$. – J Cian Oct 5 '15 at 1:45