Define the Inverse Fourier Transform (IFT) as

$f(t) = \int_{-\infty}^{{\infty}}f(\omega) \exp(i \omega t) d\omega$

I require the IFT of a decaying Gaussian

$f(\omega) = \frac{\exp(-s^2 \omega^2/2)}{i \omega + 1/\tau}$

I call it decaying because of a very similar known IFT

$g(\omega) = \frac{1}{i \omega + 1/\tau}$

$g(t) = F[g(\omega)] = H(t) \exp(-t/ \tau)$

where $H(t)$ is the (Heaviside) step function.

Please help me find the IFT of $f(\omega)$

  • $\begingroup$ Why not use the convolution theorem ? $e^{-\omega^2/2}$ is its own Fourier transform $\endgroup$ – reuns Jul 21 '17 at 19:20

You have $$\int_{\mathbb{R}}\frac{\exp[-s^{2}\omega^{2}/2+i\omega{t}]}{i\omega+1/\tau}d\omega$$ The integrand has a pole at $\omega=i/\tau$. Here I assume $\tau>0$, so the pole is in the upper half plane. You take a contour which is an infinite semi-circle in the upper half plane. The residue at the pole is $$\lim_{\omega\rightarrow{i/\tau}}(\omega-i/\tau)\frac{\exp[-s^{2}\omega^{2}/2+i\omega{t}]}{i\omega+1/\tau}=\frac{\exp[s^{2}/2\tau-{t}/\tau]}{i}$$ So the integral is $$\int_{\mathbb{R}}\frac{\exp[-s^{2}\omega^{2}/2+i\omega{t}]}{i\omega+1/\tau}d\omega=2\pi{i}\frac{\exp[s^{2}/2\tau-{t}/\tau]}{i}=2\pi\exp[s^{2}/2\tau-{t}/\tau]$$

  • $\begingroup$ Hey, thanks a lot man. I do not completely understand your proof yet, since I forgot most of my complex analysis by now, but I think there is an error. In particular, $s^2 / 2\tau$ is not a dimensionless quantity, but all arguments of the exponent have to be dimensionless $\endgroup$ – Aleksejs Fomins Jul 21 '17 at 15:07
  • 2
    $\begingroup$ @Kiryl Pesotski It should be $\exp \left(\frac{s^{2}}{2 \tau^{2}} - \frac{t}{\tau}\right)$ in the second equation. This resolves the dimension problem. $\endgroup$ – fourierwho Jul 21 '17 at 15:25
  • $\begingroup$ Ok, second problem. When one Fourier-transforms your result, one gets $\frac{\exp \biggl(\frac{s^2}{2 \tau^2}\biggr)}{i\omega + 1/\tau}$, which is not the original function $\endgroup$ – Aleksejs Fomins Jul 21 '17 at 15:42
  • $\begingroup$ I'm sorry, I made a small mistake when calculating the fourier-transform. In fact, it seems to me that the fourier transform of the result you have obtained does not converge $\endgroup$ – Aleksejs Fomins Jul 21 '17 at 15:52
  • $\begingroup$ Ok, I see the math now a bit better. Why did you assume that the integral over the semi-circle is zero? $\endgroup$ – Aleksejs Fomins Jul 21 '17 at 16:12

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.