Does the sign matter on the frequency response?
I assume you are getting the negative sign with the Fourier transform, rather than with the voltage divider approach. So, just to perform the Fourier analysis for a series RL circuit with the output taken across the resistor, the impulse response is: $$\frac{1}{\tau}e^{-t/\tau}$$ and the resultant Fourier transform is:
$$\int_{0}^{\infty}\frac{1}{\tau}e^{-t/\tau}e^{-j\omega t}dt$$

where the lower limit is zero since the impulse response is zero for \$t&lt;0\$.

Performing the integral gives:
$$\frac{1}{\tau}\left[\frac{-1}{\frac{1}{\tau}+j\omega}\large e^{-(\frac{1}{\tau}+j\omega)t}\right]_0^\infty$$

Taking the upper limit gives zero; and the lower limit gives:
$$\frac{1}{1+j\omega \tau}$$

which is positive!