I am reading a Fourier Transform definition in two places, in the first is

$$\int_{-\infty}^{\infty}f(x)\exp(-ijw)dx$$

and another is

$$\int_{-\infty}^{\infty}f(x)\exp(-2\pi ijw)dx$$

I want know Why the first is without $(2\pi)$?.

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 What are $i$ and $j$? Please check that you have transcribed what you have read correctly and completely. – Dilip Sarwate Nov 2 '12 at 17:07 Your formulas are false. For example, the exponential terms in the integrals does not depend on the x, so the formulas are meaningless, they are not describing Fourier transform. True formula for the Fourier transform is, $$F(\omega) = \int f(x) \exp(-i\omega x) dx$$ Corresponding second formula is, $$F(f) = \int f(x) \exp(-i 2 \pi f x) dx$$ You just define $\omega = 2 \pi f$, and this doesn't make difference in the general sense. – Theo Nov 2 '12 at 18:10

You will find many different expressions for the Fourier transform $$\hat f_{a,b}(\omega) = \frac{b}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x) e^{-i a x \omega}\,dx$$ with $a,b\in \mathbb{R}$.
The different Fourier transform obey the relation $$\hat f_{a,b}(\omega) = b \hat f_{1,1}(a \omega)$$ so they essentially all have the same information.
The choice of $2\pi$ is presumably to make the Fourier-Plancherel transform a unitary operator on $L_2(\mathbb R^k)$. – kahen Nov 2 '12 at 16:57
They're essentially equivalent - the only difference is a rescaling of $x$. They both lead to Fourier transforms which differ by a multiplicative constant, and so as long as the definition of the inverse Fourier transform is consistent with your choice, it doesn't matter.