FT of exp(−a|t|)u(t)

Because of $u(t)$, the signal is zero for $t<0$, and for $t>0$ we have $|t| = t$. So

$$x(t) = e^{-a|t|}u(t) = e^{-at}u(t), \quad a > 0$$

Fourier transform:

$$\begin{aligned} X(j\omega) &= \int_{0}^{\infty} e^{-at}\,e^{-j\omega t}\,dt = \int_0^\infty e^{-(a+j\omega)t}dt \\ &= \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_0^\infty = \frac{1}{a+j\omega} \end{aligned}$$

Magnitude and phase:

$$|X(j\omega)| = \frac{1}{\sqrt{a^2+\omega^2}}, \qquad \angle X(j\omega) = -\tan^{-1}\frac{\omega}{a}$$

Sketches:

 x(t)
 1 ^.
   | .
   |  ' .
   |      ' . . _ _
---+---------------> t
   0

 |X(jw)|
  1/a         *
            * | *
0.707/a  *.........*   (half-power points, w = +/- a)
      *       |       *
  *           |           *
--------------+--------------> w
        -a    0    a

 phase
 +90 | ......
 +45 |       .
   0 +--------.---------> w
     |        0 .
 -45 |            .
 -90 |              ......

The magnitude is even, the phase is odd, and the spectrum is low-pass with half-power frequency $\omega = a$.

(Note: if the signal were $e^{-a|t|}$ without $u(t)$, the answer would be $\frac{2a}{a^2+\omega^2}$, which is real and even.)