$\hat f_\lambda(x) = \hat f(x/\lambda)$ This is a small part of a larger problem I am trying to solve. This is stated as a basic property of the fourier transform. First we define for $f \in L^1(\mathbb{R}^d)$ and $\lambda \neq 0$,
$$ \hat f(x) := \int_{\mathbb{R}^d} f(t) e^{-2\pi i t x} dt 
\quad\text{and}\quad
\hat f_\lambda(x) := \lambda f(\lambda x). $$

The property is that, $\hat f_\lambda(x) = \hat f(x/\lambda)$

I don't see how to prove this, not sure where to start really.

EDIT: The property I am trying to prove is the scaling property of this wikipedia article. http://en.wikipedia.org/wiki/Fourier_transform#Basic_properties
Following the notation of the wiki article. If $h(x)=f(ax)$, then $\hat{h}(\xi)=|a|^{-1}\hat{f}\left(\frac{\xi}{a}\right)$.

Proof:

$\hat{h}(\xi)=\int h(x)e^{-2\pi i\xi x}\, dx=\int f(ax)e^{-2\pi i\xi x}\, dx=|a|^{-1}\int f(y)e^{-2\pi i\xi \left(\frac{y}{a}\right)}\, dy$.

This becomes: $|a|^{-1}\int f(y)e^{-2\pi i\left(\frac{\xi}{a}\right)y}\, dy=|a|^{-1}\hat{f}\left(\frac{\xi}{a}\right)$.

Note this was for functions taking values in $\mathbf{R}$, in higher dimensions, say $n$,  the $|a|$ gets an exponent of $-n$.  I think that is the only change.
In this case
$$\hat f(\lambda t) = \int_{\mathbb{R}^d} f(x)e^{2\pi i \lambda x t}\,dx = 
 \int_{\mathbb{R}^d} f(x/\lambda)e^{2\pi i  xt}{dx\over \lambda^d} $$ 
Note the $d$ power owing to the Jacobian of the transformation on $\mathbb{R}^d$.