Characteristic function of a random variable with absolute value in its density function

Question

If I want to find the characteristic function of $X$ with density function $f(x)=\frac {1}{2}e^{-\lvert x\rvert}$, is it true that I will get two characteristic functions depending on the value of $x$?

Answer 1

no, you won't get two different cf's:

\begin{align*} E[e^{itX}] &= \frac{1}{2}\int_{-\infty}^{\infty} e^{itx} e^{-|x|} dx\\ &= \frac{1}{2}\int_{-\infty}^{0} e^{itx} e^{x} dx + \frac{1}{2}\int_0^{\infty} e^{itx} e^{-x} dx\\ &= \frac{1}{2}\int_{-\infty}^{0} e^{x(it+1)}dx + \frac{1}{2}\int_0^{\infty} e^{x(it-1)} dx\\ &=\frac{1}{2} \left (\frac{1}{it+1} - \frac{1}{it-1} \right )\\ &= \frac{1}{2} \left(\frac{-2}{-t^2-1} \right )\\ &= \frac{1}{t^2+1} \end{align*}