Why does dimensional analysis seem to fail in determining the scaling of the Fourier transform? In his book Street-Fighting Mathematics, at pag. 7, S. Mahajan uses the technique of dimensional analysis to obtain very quickly the following result: 
$$\int_{-\infty}^{\infty}e^{-ax^2}\, dx = \frac{\text{constant}}{\sqrt{a}}.^{[1]}$$
Here a "constant" is something not depending on the parameter $a$. The trick is to observe that if $x$ has the dimension of $\verb+length+$ then $a$ must have the dimension of $\verb+length+^{-2}$. 
Now I have tried to use the same technique to derive the scaling formula for the Fourier transform:
$$\mathcal{F}[f(\lambda x)](\xi)=\frac{1}{\lambda}\widehat{f}\left(\frac{\xi}{\lambda}\right), $$
but I get the wrong result and I cannot understand why. My reasoning is the following: 


*

*Assume that $x$ has the dimensions of $\verb+length+$ and that $f(x)$ is dimensionless. 

*From the formula $$\hat{f}(\xi)=\int_{-\infty}^{\infty} f(x) e^{-i x\xi}\, dx, $$ we see that $\xi$ must have the dimensions of $\verb+length+^{-1}$ and $\widehat{f}(\xi)$ must have the dimensions of $\verb+length+$. 

*Apply the scaling $x\to \lambda x$. Then we get from point 1 that $$f(x)\to f(\lambda x)$$ and we get from point 2 that \begin{equation}\begin{array}{ccc} \xi\to \frac{\xi}{\lambda} &\text{and}& \widehat{f}(\xi)\to \lambda\widehat{f}\left(\frac{\xi}{\lambda}\right)\end{array}.\end{equation}

*Conclude that $$\tag{!!}\mathcal{F}[f(\lambda x)](\xi)=\lambda \widehat{f}\left(\frac{\xi}{\lambda}\right), $$ which is wrong.


Where am I wrong?
Thank you for reading.

[1] The precise value of the constant is $\sqrt{\pi}$. Of course this cannot be determined by dimensional analysis alone.
 A: The easy way to derive the transformation property is using the substitution $y=\lambda x$:
$$\mathcal{F}[f(\lambda x)](\xi) = \int_{-\infty}^\infty f(\lambda x)e^{-i x\xi}dx =  \int_{-\infty}^\infty f(y)e^{-ix\xi/\lambda }\frac{dx}{\lambda} = \frac{1}{\lambda}\hat{f}\left(\frac{\xi}{\lambda}\right).$$
However, since you asked about dimensional analysis, let me give you an argument based on it. Fix function $f$ and regard the Fourier transform
$$\mathcal{F}[f(\lambda x)](\xi)$$
as a function of $\lambda$ and $\xi$. Suppose $x$ has dimension of $\mathrm{length}$. Since $\lambda$ and $\xi$ appear only in combinations $\lambda x$ and $\xi x$ repectively, they both must have dimensions $\mathrm{length}^{-1}$. Since the total dimension of $\mathcal{F}[f(\lambda x)](\xi)$ is $\mathrm{length}$, it must be equal to
$$ \frac{1}{\lambda}g\left(\frac{\xi}{\lambda}\right),$$
for some (so far unknown) function $g$. Setting $\lambda=1$ reveals that $g$ is the Fourier transform of $f$.
To illustrate the point more clearly, note that, based on dimensionality only, we could have equally well said that $\mathcal{F}[f(\lambda x)](\xi)$ must be
$$ \frac{1}{\xi}h_1\left(\frac{\lambda}{\xi}\right),\quad\mathrm{or}\quad \frac{\xi}{\lambda^2}h_2\left(\frac{\xi}{\lambda}\right),\quad\ldots$$
for some $h_{1,2}$ not fixed by dimensional analysis. Setting $\lambda=1$ would then show that
$$h_1(\xi)=\frac{1}{\xi}\hat{f}\left(\frac{1}{\xi}\right),\quad h_2(\xi)=\frac{1}{\xi}\hat{f}\left(\xi\right).$$
All of the above probably assumes reasonable properties of $f$.
