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So I have the function $$f:f(t)=\frac{4}{t^2+9}$$ and I know that

\begin{align*} \mathcal{F}\{e^{-|t|}\}(\omega) &= \frac{2}{\omega^2+1}. \end{align*}

I'd gotten proof of the latter Fourier Transformation in another question thread, but whenever I try to compute the earlier function with fourier transformation, it seems very hard to do at its current form. Wolfram Alpha can do it just fine, but computing it with a calculator just won't do.

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Grak
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    Hint: Use Duality theorem of frequency and time domain. – K.K.McDonald Feb 21 '17 at 11:28
  • Does switching from f(t) to f(w) entail something other than simply switching t into w, or have I missed something in regards from transforming a function from time-level to frequency-level? – Grak Feb 21 '17 at 11:38
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    There is this relation: $$f(t) \rightarrow F(\omega)$$ $$F(t) \rightarrow 2\pi f(-\omega)$$. And proof of it is quite simple. – K.K.McDonald Feb 21 '17 at 12:07
  • Is it a "Fourier transformation" if I make $F(t)$, or do I have to process it further down the line? – Grak Feb 21 '17 at 12:55

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Let $$ g(t) := \frac{2}{t^2+1} $$ In an answer to a previous question of yours, it was shown, using the inversion theorem, that $$ \mathcal{F}\left\{\frac{2}{t^2+1}\right\}(\omega) = 2\pi e^{-|\omega|} $$ Now note that $f(t) = \frac{2}{9} g(t/3)$ and recall the following result.

Proposition: If $f(t) = g(at)$ where $a \in \Bbb{R}\backslash\{0\}$ then $$\mathcal{F}\{f\}(\omega) = \frac{1}{|a|}\mathcal{F}\{g\}(\omega/a)$$

Thus \begin{align} \mathcal{F}\{f\}(\omega) &= \frac{2}{9}\cdot\left(3\cdot2\pi e^{-|3\omega|}\right)\\ &=\frac{4\pi}{3}e^{-3|\omega|} \end{align}

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