I need help to calculate the Fourier transform of this function:
$$\frac1{t^2+2t+2}$$
Thanks!
We know that
$$\mathcal{F}(e^{-a|x|})=\frac{2a}{w^2+a^2}$$
Therefore
$$\frac{2a}{x^2+a^2}\overset{\mathcal{F}}{\longrightarrow}2\pi e^{-a|w|}$$
Also
$$f(x-a) \overset{\mathcal{F}}{\longrightarrow} e^{-iaw}\mathcal{F(w)}$$
From the last formula and $a=1$ we have that
$$\mathcal{F}\left(\frac{1}{t^2+2t+2}\right)=\mathcal{F}\left(\frac{1}{(t+1)^2+1}\right)=e^{iaw}\mathcal{F}\left(\frac{1}{t^2+1}\right)=(e^{iaw})(\pi e^{-|w|})$$
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We know that
$$\mathcal{F}(e^{-a|x|})=\frac{2a}{w^2+a^2}$$
Therefore
$$\frac{2a}{x^2+a^2}\overset{\mathcal{F}}{\longrightarrow}2\pi e^{-a|w|}$$
Also
$$f(x-a) \overset{\mathcal{F}}{\longrightarrow} e^{-iaw}\mathcal{F(w)}$$
From the last formula and $a=1$ we have that
$$\mathcal{F}\left(\frac{1}{t^2+2t+2}\right)=\mathcal{F}\left(\frac{1}{(t+1)^2+1}\right)=e^{iaw}\mathcal{F}\left(\frac{1}{t^2+1}\right)=(e^{iaw})(\pi e^{-|w|})$$