# What's the difference between these two definitions of Fourier transformation of measure?

Let $\mu$ be a probability measure on $\Bbb R^d$. I met the following two definitions of Fourier transformation of $\mu$ in the textbook:

1. $\displaystyle \widehat{\mu}(\xi)=\int e^{i\xi\cdot x}d\mu(x)$;
2. $\displaystyle \widehat{\mu}(\xi)=\int e^{-2\pi i\xi\cdot x}d\mu(x)$.

My question is: what's the difference between these two definitions of Fourier transformation of measure?

• Do you mean $\widehat{\mu}(\xi)=\int e^{-i\xi\cdot x}d\mu(x)$ in the first case? – Sayan Dec 5 '15 at 14:27
• @Sayan: Not necessarilly. That one is yet another convention... – Alex M. Dec 5 '15 at 14:39
• No, Sayan. The first case is just as I gave. – user47280 Dec 5 '15 at 14:47
• A nice feature of definition 2 is that it removes the need for a prefactor on the inverse transformation. On the other hand every derivative pops out a factor of $-2\pi i\xi$ which can be annoying when working with differential equations. – Spencer Aug 15 '16 at 18:59