I'm asked to find Fourier transform of $ f(t) = \frac{1}{t^2 + 2t + 2} $. I have searched for steps for solving this kind of problem (given a inverse polynomial) but couldn't find any. Should I factorize it to $ \frac{1}{(t+1-i)(t+1+i)} $ or? The given correct answer is $ F(w) = \pi e^{iw-|w|} $.

If there is any similar example with solution, feel free to drop link and I can study the steps.

  • $\begingroup$ Try $$t^{2} + 2t + 2 = (t+1)^{2} + 1$$ change variables to $u = t+1$ and then use this. However, if you use a transform with a different scaling (which I think you do), your solution will be slightly different to the one used in the link example. $\endgroup$
    – mattos
    Feb 10, 2017 at 1:05
  • $\begingroup$ @Mattos Thanks, didn't see that! I got similar answer to the link you posted, by using another formula $ f(t) = e^{-a|t|} $ transform to $ F(w) = \frac{2a}{a^2 + w^2}, a = 1 $. So if I substitute $ u = t+ 1 $, then I get $ \pi e^{-|-w|} = \pi e^{-|w|} $. From there I'm not sure how to continue, or substitute back the $t$. $\endgroup$ Feb 10, 2017 at 1:58
  • $\begingroup$ What transform are you using? See here for the different types. If you use the approach I used in the other linked answer, you get $$F(\omega) = \pi e^{-2 \pi( i \omega - \lvert {\omega} \rvert )}$$ $\endgroup$
    – mattos
    Feb 10, 2017 at 5:39
  • $\begingroup$ @Mattos I went with this transform link at page 3, the 2nd row. If there is a better approach, please let me know which one I should use. $\endgroup$ Feb 10, 2017 at 11:14
  • $\begingroup$ They are all exactly the same, just a change of variables will get you from one to the other. Did you manage to get a solution? $\endgroup$
    – mattos
    Feb 10, 2017 at 13:35

1 Answer 1


Using the Fourier transform of $e^{- \alpha \lvert t \rvert}$, we find

\begin{align} F(\omega) &= \frac{2 \alpha}{\alpha^{2} + \omega^{2}} \end{align}


\begin{align} e^{- \alpha \lvert t \rvert} &= \frac{1}{2 \pi} \int_{\mathbb{R}} \frac{2 \alpha}{\alpha^{2} + \omega^{2}} e^{i \omega t} d \omega \\ \implies \pi e^{- \alpha \lvert t \rvert} &= \int_{\mathbb{R}} \frac{ \alpha}{\alpha^{2} + \omega^{2}} e^{i \omega t} d \omega \\ \end{align}

Now, set $\alpha = 1$ and map $t \mapsto -t$

\begin{align} \implies \pi e^{- \lvert -t \rvert} &= \pi e^{- \lvert t \rvert} \\ &= \int_{\mathbb{R}} \frac{ 1}{1 + \omega^{2}} e^{- i \omega t} d \omega \\ \end{align}

Now map $\omega \mapsto \omega + 1$

\begin{align} \implies \pi e^{- \lvert t \rvert} &= \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i (\omega + 1) t} d \omega \\ &= e^{-i t} \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i \omega t} d \omega \\ \implies \pi e^{- \lvert t \rvert + i t} &= \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i \omega t} d \omega \end{align}

hence, by the Fourier inversion formula

\begin{align} F(\omega) &= \int_{\mathbb{R}} \frac{ 1}{1 + (t + 1)^{2}} e^{- i \omega t} dt \\ &= \pi e^{i \omega - \lvert \omega \rvert} \end{align}

  • $\begingroup$ Thanks, after studying the solution, it helped me to sort out the problem! $\endgroup$ Feb 11, 2017 at 17:20

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