Show the solution for $\mathcal{F}(e^{-\left | t \right |})$

Question

I'm trying to show that $$\mathcal{F}(e^{-\left | t \right |}) = \frac{2}{\sqrt{2\pi}(1 + w^2)}$$ Knowing that the Fourier transform is in the form $$\mathcal{F}(f(x)) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)e^{-iwt}dt$$ Someone can help me?

Answer 1

The integral splits into $$ \int_0^{\infty} e^{-t}e^{-iwt} \, dt + \int_{-\infty}^{0} e^{t} e^{-iwt} \, dt. $$ Changing variables in the second integral, you can rewrite this as $$ 2\int_0^{\infty} e^{-t} \, \cos{wt} \, dt. $$ Now integrate by parts twice, and you find $$ \int_0^{\infty} e^{-t} \, \cos{wt} \, dt = 1-w^2 \int_0^{\infty} e^{-t} \, \cos{wt} \, dt, $$ which you can rearrange and then multiply by the other factors to get the result.

Answer 2

Just to be clear with the post below, the reason to split the integral as such is because: $$|t|= \left\{ \begin{array}{lr} -t & : t \le 0\\ t & : t > 0 \end{array} \right.\\ $$

$$f(t)=e^{-|t|}= \left\{ \begin{array}{lr} e^{-(-t)} & : t \le 0\\ e^{-(t)} & : t > 0 \end{array} \right.\\ $$

$$f(t)= \left\{ \begin{array}{lr} e^{t} & : t \le 0\\ e^{-t} & : t > 0 \end{array} \right.\\ $$