How can I fourier transform this function, with a special character inside

Question

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.

Answer 1

\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 $t \in \Bbb{R}$.