# Fourier transform of the signal $x(t)=e^{-a|t|}$ , a>0

I having a problem doing the Fourier transform of the signal $x(t)=e^{-a|x|}$, when a > 0.

When evaluating $\int_{-\infty}^{0}e^{at}e^{-jwt}dt$ , it seems to diverge, but somehow it converges and I can't see why, can someone help me with this one.

Because $e^{t(a-jw)}=e^{at-jwt}$ , and when we evaluate, $at$ goes to $\frac 1{\infty}=0$, but $-jwt$ goes to ${\infty}$, and we will have $0*"{\infty}"$

• $\forall t,w \in \mathbb{R}, \left| e^{-jwt} \right| = 1$, so there is no problem. – nicomezi Jun 8 '18 at 5:39
• You can find the FT by separating real and imaginary parts and using standard Calculus formulas – Kavi Rama Murthy Jun 8 '18 at 6:24

$\int_{-\infty}^0 e^{-a|t|} e^{-i \omega t} dt = \int_0^{\infty} e^{-as} e^{i \omega s} ds = {1 \over a- i \omega}$.
Hence $({\cal F} x)(\omega)={1 \over a- i \omega}+{1 \over a+ i \omega} = {2a \over a^2+\omega^2}$ which matches that in https://en.wikipedia.org/wiki/Fourier_transform#Square-integrable_functions.
• The question is about convergence and the $0\cdot\infty$ issue. The proper answer was given by nicomezi in a comment. – Yves Daoust Jun 8 '18 at 7:48
• @YvesDaoust: I am well able to read. By couching the problem in terms of $[0, \infty)$ my intent was that the OP would see why it converges. – copper.hat Jun 8 '18 at 23:01