What is the relationship between different definitions of Fourier transform? I always see various definitions of Fourier transform. A standard form is:
$$\hat{f}(\xi)=\int_{\mathbb{R}^d}f(x)e^{-2\pi ix\cdot\xi}dx$$
and its attached inversion is
$$f(x)=\int_{\mathbb{R}^d}\hat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi$$
Another form is like this:
$$\hat{f}(\xi)=\int_{\mathbb{R}^d}f(x)e^{-ix\cdot\xi}dx$$
and the inversion formula is
$$f(x)=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d}\hat{f}(\xi)e^{ix\cdot\xi}d\xi$$
I believe they are actually the same and I try to find their relationship.
An article on ProofWiki says

There exist several slightly different definitions of the Fourier transform which are commonly used; they differ in the choice of the constant 2π inside the exponential and/or a multiplicative constant before the integral. 
  Their properties are essentially the same, and by a simple change of variable one can always translate statements using one of the definitions into statements using another one. 

So I tried change of variable:
$$ \int_{\mathbb{R}^d}f(x)e^{-ix\cdot\xi}dx=(2\pi)^d\int_{\mathbb{R}^d}f(2\pi x)e^{-2\pi ix\cdot\xi}dx$$
But then since the variable in $f$ is $2\pi ix$ rather than $x$, I don't know how to deal with it.
Can you please help? Thank you.
EDIT: According to James Edward Lewis, the change of variable should be $2\pi x$. I revised this.
 A: The difference between the first and the second form is in the definition of $\xi$ (so you should change the name of the variables correspondingly). Let me introduce the new notation $\tilde\xi$ and $\tilde f(\tilde \xi)$ such that the confusion is (hopefully) lifted.
You want to proof that 
$$ \hat{f}(\xi)=\int_{\mathbb{R}^d}f(x)e^{-2\pi ix\cdot\xi}dx \qquad 
f(x)=\int_{\mathbb{R}^d}\hat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi$$
is equivalent to
$$ \tilde{f}(\tilde\xi)=\int_{\mathbb{R}^d}f(x)e^{-ix\cdot\tilde\xi}dx \qquad
f(x)=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d}\tilde{f}(\tilde\xi)e^{ix\cdot\tilde\xi}d\tilde\xi.$$
Let's take the first form as given and try to derive the second form. The Fourier transform is by definition the same with $\tilde\xi = 2\pi \xi$ and $\tilde f(\tilde \xi) = \hat{f} (\xi)$. For the inversion formula, we have to work a bit harder (and use substitution).
$$f(x)=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d}\tilde{f}(\tilde\xi)e^{ix\cdot\tilde\xi}d\tilde\xi =
\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d}\tilde{f}(\tilde\xi)e^{2\pi ix\cdot\xi}d(2 \pi \xi)
= \int_{\mathbb{R}^d}\hat{f}(\xi)e^{2\pi ix\cdot\tilde\xi}d\xi.$$
A: The correct change of variable is x=2pi*X, and in the end you actually will be using a different function; to put it another way, the first form of Fourier transform on a function f is the same as the second form on its pullback by the linear map consisting of multiplication by 2pi, divided by (2pi)^d.
The properties are essentially the same, but the transforms are not exactly the same; the trade-off is between simplicity and asymmetry of inversion.
A: Sorry Fabian, but i think James Edward Lewis II is absolutely right with his Statement, that $\tilde f(\tilde \xi) \neq \hat{f} (\xi)$ and there is an easy way of getting all definition of the Fourier transform:


*

*My favourite definition: $$  \hat f_1(k_1)=\int_{\mathbb{R}}f_1(x_1)e^{-i2\pi k_1 x_1}dx_1 \hspace{10mm} f_1(x_1)=\int_{\mathbb{R}}\hat f_1(k_1)e^{-i2\pi k_1 x_1}dk_1 \hspace{10mm} (1)$$

*Let $k_2=2\pi k_1$, so we can define for this new variable also the corresponding $\hat f_2(k_2)$, which is also a Fourier-transform of $f_1(x_1)$ with just other units, "radians/meter" if you like, in contrast to $\hat f_1(k_1)$, which has "circles/meter". The function $\hat f_2(k_2)$ is just a bit stretched, but with invariant area, therefore:
$$ \hat f_2(k_2)|dk_2|=\hat f_1(k_1)|dk_1| \hspace{5mm} => \hspace{5mm} \hat f_2(k_2) = \hat f_1(k_1) \left|\frac{dk_1}{dk_2}\right|=\frac{1}{2\pi}\hat f_1(k_1)$$
$$=> \hspace{5mm}\hat f_2(k_2)=\frac{1}{2\pi}\int_{\mathbb{R}}f_1(x_1)e^{-ik_2 x_1}dx_1 \hspace{10mm}f_1(x_1)=\int_{\mathbb{R}}\hat f_2(k_2)e^{-ik_2 x_1}dk_2 \hspace{5mm}(2)$$

*Let $x_3 = 2\pi x_1$, and with the same argument, we have a new $f_3(x_3)$ with
$$ f_3(x_3) = f_1(x_1) \left|\frac{dx_1}{dx_3}\right|=\frac{1}{2\pi} f_1(x_1)$$
$$=> \hspace{5mm}\hat f_1(k_1)=\int_{\mathbb{R}}f_3(x_3)e^{-ik_1 x_3}dx_3 \hspace{5mm}f_3(x_3)=\frac{1}{2\pi}\int_{\mathbb{R}}\hat f_1(k_1)e^{-ik_1 x_3}dk_1 \hspace{5mm}(3)$$

*Let $x_4 = \sqrt{2\pi} x_1$ and $k_4 = \sqrt{2\pi} k_1$, therefore we have a new function $f_4(x_4)$ and a new function $\hat f_4(k_4)$ with 
$$ f_4(x_4) = f_1(x_1) \left|\frac{dx_1}{dx_4}\right|=\frac{1}{\sqrt{2\pi}} f_1(x_1)$$
$$ \hat f_4(k_4) = \hat f_1(k_1) \left|\frac{dk_1}{dk_4}\right|=\frac{1}{\sqrt{2\pi}} \hat f_1(k_1)$$
$$=> \hspace{5mm}\hat f_4(k_4)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f_4(x_4)e^{-ik_4 x_4}dx_4 \hspace{5mm}f_4(x_4)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat f_4(k_4)e^{-ik_4 x_4}dk_4 \hspace{5mm}(4)$$
The first definition of the Fourier transform is much more symmetric and pretty, and therefore i want to show you a little math joke:
$$ \hat f(k)=\int_{\mathbb{R}}f(x)e^{-i2\pi k x}dx \stackrel{?}{=} \int_{\mathbb{R}}f(x)\left(e^{-i2\pi}\right)^{k x}dx = \int_{\mathbb{R}}f(x)1^{k x}dx = \hat f(0)$$
Does anyone get it?
