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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 $f(\omega)$

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  • $\begingroup$ Why not use the convolution theorem ? $e^{-\omega^2/2}$ is its own Fourier transform $\endgroup$ – reuns Jul 21 '17 at 19:20
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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 $$\int_{\mathbb{R}}\frac{\exp[-s^{2}\omega^{2}/2+i\omega{t}]}{i\omega+1/\tau}d\omega=2\pi{i}\frac{\exp[s^{2}/2\tau-{t}/\tau]}{i}=2\pi\exp[s^{2}/2\tau-{t}/\tau]$$

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  • $\begingroup$ 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
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    $\begingroup$ @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

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