1
$\begingroup$

Given Fourier's Inversion theorem: $$f(t)=\displaystyle\frac{1}{2\pi}\int_{\omega =-\infty}^\infty e^{i \omega t} \, \mathrm{d}\omega \int_{u=-\infty}^\infty f(u) e^{-i\omega u} \,\mathrm{d}u \tag{1}$$ and the Fourier transform of $f(t)$: $$\displaystyle\mathcal{F}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{t=-\infty}^{\infty}\color{red}{f(t)}e^{-i\omega t}\mathrm{d}t\tag{A}$$

and the Inverse Fourier transform of $\mathcal{F}(\omega)$:

$$\displaystyle \color{red}{f(t)}=\color{blue}{\frac{1}{\sqrt{2\pi}}\int_{\omega=-\infty}^{\infty}\mathcal{F}(\omega)e^{i\omega t}\mathrm{d}\omega}\tag{B}$$

In trying to understand Fourier pairs I substitute $(\mathrm{B})$ into $(\mathrm{A})$ by replacing the part marked $\color{red}{\mathrm{red}}$ in $(\mathrm{A})$ with the expression $(\mathrm{B})$ which gives

$$\displaystyle\mathcal{F}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{t=-\infty}^{\infty}\color{blue}{\frac{1}{\sqrt{2\pi}}\int_{\omega=-\infty}^{\infty}\mathcal{F}(\omega)e^{i\omega t}\mathrm{d}\omega}e^{-i\omega t}\mathrm{d}t$$

$$\implies \mathcal{F}(\omega)=\color{#180}{\displaystyle\frac{1}{2\pi}\int_{t=-\infty}^{\infty} \int_{\omega=-\infty}^{\infty}\mathcal{F}(\omega)\mathrm{d}\omega \mathrm{d}t}\tag{C}$$

Now unless I'm mistaken the RHS of $(\mathrm{C})$ marked $\color{#180}{\mathrm{green}}$ should reduce to $\mathcal{F}(\omega)$ on the LHS of $(\mathrm{C})$.

What steps do I need to take to show this?

Thank you.

$\endgroup$
3
$\begingroup$

You should be slightly more careful with your notation. We have the Fourier transform $$ \hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{-i\omega x} \,dx $$ and Fourier inversion $$ f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \hat{f}(\xi) e^{i\xi x} \,d\xi. $$ You want to insert the first equation into the second: $$ \hat{f}(\omega) =\frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \hat{f}(\xi) e^{-i\omega x} e^{i\xi x} \,d\xi\,dx = \frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \hat{f}(\xi) e^{i(\xi - \omega)x} \,d\xi\,dx $$ The thing to realize is that the $\omega$ appearing in the argument of $\hat{f}$ on the left is not the same as the one that appears in the definition of the Fourier inversion (your Equation B). By using different letters for the dummy integration variables, you see that the exponentials do not cancel, and the correct version of your equation (C) is the one I have above. At this step there is not much else that can be simplified. Rather than thinking of simplifying your green equation (or, rather, the corrected version I propose), think of it as a result.

$\endgroup$
  • $\begingroup$ Thanks for your answer, if the Fourier transform is a transformation of $f(t)$ then the Fourier inversion is a transformation of ...? $\endgroup$ – BLAZE Oct 14 '15 at 6:54
  • $\begingroup$ Well, it is (almost the same) transformation applied to $\hat{f}(\omega)$, namely it is a transform of the Fourier transform of $f$. I'm not sure that this is mathematically meaningful though. $\endgroup$ – Andrew Oct 14 '15 at 7:03
  • $\begingroup$ I'm just trying to understand why you said "You should be slightly more careful with your notation."? $\endgroup$ – BLAZE Oct 14 '15 at 9:03
  • $\begingroup$ @Andrew In your last integral the $x$-integration leads to the delta function $δ(ξ-ω)$ and you get the left hand side back. $\endgroup$ – Urgje Oct 14 '15 at 12:48
  • 1
    $\begingroup$ @BLAZE they arent the same for the reason I described above. One $\omega$ appears as an input variable, one appears as a dummy integration variable. These are literally two different things. $\endgroup$ – Andrew Oct 18 '15 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.