I have been unable to see why the $1 / (2 \pi)$ appears in Fourier transform. Would you please justify it to me?
Problem: Let $$ f(x) = \int_{-\infty}^{\infty} \mathrm{d} k \, e^{ikx} \tilde{f}(k) \, . $$ Why does it hold that $$ \tilde{f}(k) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \mathrm{d} x \, e^{-ikx} \tilde{f}(k) \, ? $$ Especially, where does the $1 / (2 \pi)$ come from?
I think I understand how the thing should behave for functions on $x \in [0,P]$. Provided that $$ f(x) = \sum_k \tilde{f}_k e^{ikx} \, $$ for $k \in \{ \frac{2 \pi n}{P} \, \mathrm{for} \, n \in Z \}$, then $$ \int_0^P \mathrm{d} x \, e^{-ikx} f(x) = \sum_q \tilde{f}_q \int_0^P \mathrm{d} x \, e^{ix(k-q)} \, . $$ $$ \int_0^P \mathrm{d} x \, e^{ix(k-q)} = \frac{1}{i (k-q)}{e^{iP(k-q)} - 1} = P \delta_{kq} \, , $$ Therefore $$ \int_0^P \mathrm{d} x \, e^{-ikx} f(x) = P \tilde{f}_k \, . $$ To have everything properly normalized, I have $$ \tilde{f}_k = \frac{1}{P} \int_0^P \mathrm{d} x \, e^{-ikx} f(x) \, . $$
Moving to $x \in R$, the $2 \pi$ seems to play the same role as $P$, but I do not see why.
Thanks a lot.