Characteristic function of $p(x) = \frac{1}{2} e^{-|x|}$, $-\infty < x < \infty$

Question

Let X denote a real-valued random variable with an absolutely continuous distribution with density function $p(x) = \frac{1}{2} e^{-|x|}$, $-\infty < x < \infty$.

Find the characteristic function.

Attempt:

I know that the setup should look something like this:

$$\frac{1}{2} \int_{-\infty}^\infty e^{itx} e^{-|x|} \;dx $$

Answer 1

The characteristic function is

$$\begin{align} \frac12 \int_{-\infty}^{\infty} e^{-|x|} e^{i t x} ~dx &= \frac12 \int_{-\infty}^0 e^{x+i t x} ~dx + \frac12 \int_0^{\infty} e^{-x + i t x} ~dx\\ &= \frac12 \int_0^{\infty} e^{-(1+i t) x} ~dx + \frac12 \int_0^{\infty} e^{-(1-i t) x} ~dx\\ &= \frac12 \frac{1}{1+i t} + \frac12 \frac{1}{1-i t}\\ &= \frac{1}{1+t^2}\end{align}$$

Answer 2

Note that $(a+bi)+(a-bi)=2a$ so by splitting the integral at $0$ $$ \frac12\int_{-\infty}^\infty e^{itx}e^{-|x|}dx = \int_0^\infty \mathfrak{Re}(e^{itx}e^{-x})dx $$ $$ =\int_0^\infty \cos(tx)e^{-x}dx $$ $$ =e^{-x}\frac{t\sin(tx)-\cos(tx)}{t^2+1}\biggr|_{x=0}^{\infty} $$ $$ =\frac{1}{1+t^2}. $$ I will leave the calculation of the integral (by parts twice) to you.