# Fourier antitransform using scaling property?

I'm trying to calculate the antitransform of:

$$\frac{1}{2\cdot(1+5w)^2}$$

Now I know the antitransform of $$\frac{1}{(1+5w)^2} = t \cdot e^{-5t} u(t)$$

But in this case I got that divided by 2. I assumed I had to use the scaling property which says:

$$F[f(ax)] = \frac{1}{|a|} \hat{f}(\frac{w}{a})$$

Now I'm not really sure how to apply this. Could anyone help?

• Scaling property only applies is you multiply the time/frequency variable. In this case, you only need to multiply the inverse with 1/2. – Hilmar Dec 12 '19 at 5:04

If $$h(t)$$ is the inverse Fourier transform of $$H(\omega)$$, then by linearity the inverse Fourier transform of $$aH(\omega)$$ is simply $$ah(t)$$. This has nothing to do with the scaling property you mentioned, because the latter refers to the scaling of the argument of the function.