First consider the function $f(x)=e^{iax}$, then its fourier transform is given by $F(k)=2\pi \delta(k-a)$ where $\delta$ is the Dirac-Distribution or an 'Impulse'.
Now consider $g(x)=e^{i (as) x}$ for some constant $s>0$,
applying the fourier transform like above would give $G(x)=2\pi \delta(k-as)$
applying the scaling property $g(x)=f(x\cdot s)$ would give $G(k)=\frac{1}{|s|} F(\frac{k}{s})=\frac{2\pi}{|s|}\delta(\frac{k}{s}-a)=\frac{2\pi}{|s|}\delta(k-as)$
how do I compensate for missing factor $\frac{1}{s}$
Any help would be greatly appreciated
$$\delta\left(\frac{k}{s}-a\right)=\delta\left(\frac{1}{s}(k-as)\right)=|s|\delta(k-as) $$
because $$\delta(ax)=\frac{1}{|a|}\delta(x)$$