Scaled Fourier transform

Question

if I define the transform ($\epsilon \ll 1$)

$$\hat f(\omega) = \int f(t)e^{\frac{i\omega t}{\epsilon^2}}dt,$$

is the inverse transform

$$f(t) = \frac{1}{2\pi}\int \hat f(\omega)e^{-\frac{i\omega t}{\epsilon^2}}d\omega$$

Answer 1

Let $$\hat f_\epsilon(\omega) = \int f(t)e^{\frac{i\omega t}{\epsilon^2}}dt.$$

Then $\hat f_\epsilon(\omega) = \hat f_1(\omega/\epsilon^2),$ with $\hat f_1$ being the unscaled transform. Thus, $\hat f_1(\omega) = \hat f_\epsilon(\epsilon^2\omega).$

Using the inverse transform we get $$ f(t) = \frac{1}{2\pi} \int \hat f_1(\omega) e^{-i\omega t} \, d\omega = \frac{1}{2\pi} \int \hat f_\epsilon(\epsilon^2\omega) e^{-i\omega t} \, d\omega. $$ Making the substitution $\omega'=\epsilon^2\omega$ we then get $$ f(t) = \frac{1}{2\pi} \int \hat f_\epsilon(\omega') e^{-i\omega' t/\epsilon^2} \, d\omega'/\epsilon^2 = \frac{1}{2\pi\epsilon^2} \int \hat f_\epsilon(\omega') e^{-i\omega' t/\epsilon^2} \, d\omega'. $$ Thus your inversion formula is missing a factor $\frac{1}{\epsilon^2}.$