'Obtain' the Fourier transform

Question

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?

Answer 1

Yes, write out the Fourier transform and try to evaluate the integral.

You can split the integral into two parts, one from $-\infty$ to $0$, and one from $0$ to $\infty$. Now think about what the sign of $t$ should be in each of those cases to allow it be equal to the absolute value and realize that you don't need the absolute value anymore (if you select the proper sign for $t$, that is).

Try it, and if it does not work out, let me know in the comments, then I will give a more explicit answer.

Answer 2

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*}

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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 $$ \mathcal{F}\left(e^{-a |t|}\right)(x) =\frac{1}{\pi}\frac{ a}{ a^2+ x^2} \qquad \forall a \in \mathbb{R}_+^*. $$