'Obtain' the Fourier transform If $g(t) = e^{-a|t|}$ and a is a real positive constant, obtain the fourier transform. 
I'm a bit unsure what this is asking. I can write out the expression for the fourier transform. Should I stop there? Or do I do the integration as well? If I do the integration, how do I deal with the absolute value symbol? And the fact that the limits are positive and negative infinity?

We note that $\displaystyle \mathcal{F}\left(f(t)\right)(x)=\int_{-\infty}^{+\infty}f(t)e^{-2i\pi x t}\text{ d}t.$
 \begin{eqnarray*}
\mathcal{F}\left(e^{-a |t|}\right)(x) &= & \int_{-\infty}^0 e^{ at}e^{-2i\pi x t} dt+\int_{0}^{+\infty} e^{-at}e^{-2i\pi x t} dt \\
                                     &= &\left[\frac{e^{ at-2i\pi x t}}{ a-2i\pi x}\right]_{-\infty}^0+\left[\frac{e^{-at-2i\pi x t}}{- a-2i\pi x}\right]_{0}^{+\infty}\\
                                     &= &\frac{1}{ a-2i\pi x}- \frac{1}{- a-2i\pi x}\\
                                     &= & \frac{2 a}{a^2+(2\pi x)^2}\\
                                     &= &\frac{2 a}{ a^2+ 4 \pi^2 x^2}
                                     \end{eqnarray*}

If we assume the $\mathcal{F}$ourier transform can be writen as:
$$\displaystyle \mathcal{F}\left(f(t)\right)(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(t)e^{-i x t}\text{ d}t.$$
We can find  $$
\mathcal{F}\left(e^{-a |t|}\right)(x) =\frac{1}{\pi}\frac{ a}{ a^2+  x^2}
                                    \qquad \forall a \in \mathbb{R}_+^*. $$
