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

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.
