Change of variable $2\pi$ in the Fourier Transform Introduction:
Apparently the coefficient $\frac{1}{2\pi}$ in the Fourier transform formulas (the FT and the IFT) can be either:

*

*in front of the FT, and not in front of the IFT

*in front of the IFT, and not in front of the FT

*in front of both (but root-squared)

As we have the choice (apparently), let's consider the first case.
the problem:
I'd like to know if the form $$ \hat f (x) = \frac{1}{2\pi} \int f(t) e^{-ixt} dt \tag{1}$$
and the form $$ \hat f (x) = \int f(t) e^{-i2\pi xt} dt \tag{2} $$
are equivalent?
To go from $(1)$ to $(2)$ I tried the change of variable $u=\frac{t}{2\pi}$ and it yielded to:
$$\hat f (x) = \int f(2\pi t) e^{-i2\pi xt} dt \tag{3}$$
I can't see the correspondence between $(2)$ and $(3)$ because when the integration is performed, at $t=t_0$ the term $f(2\pi t_0)$ and $f(t_0)$ are different but they both are multiplied by the same factor $e^{-i2\pi xt_0}$ so the result of the integration can't be the same ?
 A: In your equations $(1)$ and $(2)$, the right-hand side terms labeled $\hat f$ are not the same.

We begin with an $L^2$ function $f$ and define its Fourier transform.  If for example, we define the Fourier transform as
$$F(x)=\int_{-\infty}^\infty f(t)e^{-i2\pi xt}\,dt\tag1$$
then we can recover $f$ from the inverse Fourier transform as given by
$$f(t)=\int_{-\infty}^\infty F(x)e^{i2\pi tx}\,dx\tag 2$$
Note that $(1)$ and $(2)$ define the Fourier transform pair $f$ and $F$.
As an example, suppose $f(t)=1$ for $|t|<1$ and $0$ elsewhere.  Then, $F(x)=\frac{\sin(2\pi x)}{\pi x}$

If we define another transform as
$$G(x)=\int_{-\infty}^\infty f(t) e^{-ixt}\,dt\tag3$$
then we recover $f$ from the inverse transform as
$$f(t)=\frac1{2\pi}\int_{-\infty}^\infty G(x)e^{itx}\,dx\tag4$$
Note that $(3)$ and $(4)$ comprise another transform pair $f$ and $G$.
As an example, suppose $f(t) =1$ for $|t|\le 1$ and $0$ elsewhere.  Then, $G(x)=\frac{\sin(x)}{\pi x}$

If we wish to find the relationship between $F$ and $G$, we simply note that
$$\frac1{2\pi}\int_{-\infty}^\infty G(x)e^{itx}\,dx=\int_{-\infty}^\infty G(2\pi x)e^{i2\pi tx}\,dx$$
from which uniqueness of the Fourier transform implies that $F(x)=G(2\pi x)$ or equivalently, $G(x)=F(x/2\pi)$.
