Why is the absolute value needed with the scaling property of fourier tranforms? I understand how to prove the scaling property of Fourier Transforms, except the use of the absolute value:
If I transform $f(at)$ then I get $F\{f(at)\}(w) = \int f(at) e^{-jwt} dt$ where I can substitute $u = at$ and thus $du = a dt$ (and $\frac{du}{a} = dt$) which gives me:
$ \int f(u) e^{-j\frac{w}{a}u} \frac{du}{a} = \frac{1}{a} \int f(u) e^{-j\frac{w}{a}u} du = \frac{1}{a} F \{f(u)\}(\frac{w}{a}) $
But, according to various references, it should be $ \frac{1}{|a|} F \{f(u)\}(\frac{w}{a}) $ and I don't understand WHY or HOW I get/need the absolute value here?

Think about the range of the variable $t$ in the integral that gives the transform. How do the 'endpoints' of this improper integral transform under $t\to at$? Can you see how this depends on the sign of $a$?
