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picture of function

This is the function I'm dealing with, and I'm wondering what the special characteristics are of one such function, especially when observing the special function name, that would normally be "f".

I'm to fourier transform it, and it shouldn't be too hard except that I think that I have to do something to the function before transforming it because of the unknown special characteristics it might have.

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    I don't understand your question. Do you want to compute the Fourier transform of $\displaystyle\frac{2}{\omega^2+1}$ or do you want to compute the Fourier transform of $e^{-|t|}$ and verify that it is indeed $\displaystyle\frac{2}{\omega^2+1}$ ? What is the "special function name" you're talking about? Are you talking about $\mathcal{F}\{e^{-|t|}\}(\omega)$ ? The symbol $\mathcal{F}$ is used to indicate that this is the Fourier transform of what follows. In this case $\mathcal{F}\{e^{-|t|}\}(\omega)$ is a notation for the Fourier transform (function of $\omega$) of the function $e^{-|t|}$. – 2017-02-21
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    Your take on what the function symbol means is probably right. So as long as it is right, does that mean that I'm meant to prove that fourier transform of $e^{-|t|}$ is as written? – 2017-02-21
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    I can only guess so. That should be a very easy proof. – 2017-02-21

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\begin{align*} \mathcal{F}\{e^{-|t|}\}(\omega) &= \int_{-\infty}^{\infty} e^{-|t|} e^{-i t \omega}\,dt \\ &= \int_{-\infty}^{0} e^{t(1-i\omega)} \,dt + \int_0^{\infty} e^{-t(1+i\omega)} \,dt \\ &= \frac{1}{1-i\omega} + \frac{1}{1+i\omega} \\ &= \frac{2}{\omega^2+1} \end{align*} Is this detailed enough for you?

If what you want is the Fourier transform of $\displaystyle\frac{2}{\omega^2+1}$ then the easiest way to go is certainly to apply the

Inversion theorem: Let $f \in L^1(\mathbb{R})$ such that $\mathcal{F}\{f\} \in L^1(\mathbb{R})$. Then, at every point $t$ where $f$ is continuous, we have $$ \mathcal{F}\{\mathcal{F}\{f\}\}(t) = 2\pi f(-t) $$

Hence \begin{align} \mathcal{F}\{\mathcal{F}\{e^{-|t|}\}\}(t) &= \mathcal{F}\left\{\frac{2}{\omega^2+1}\right\}(t) \\ &= 2\pi e^{-|-t|} \\ &= 2\pi e^{-|t|} \end{align} for all $t \in \Bbb{R}$.

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    Very much so, thanks – 2017-02-21