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

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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?

## marked as duplicate by Matt L., jojek♦, Peter K.♦Oct 5 '15 at 20:11

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