Characteristic function of Laplace distribution I'm trying to derive the characteristic function for the Laplace distribution with density $$\frac{1}{2}\exp\{-|x|\}$$
My attempt:
$$\frac{1}{2}\int_{\Omega}e^{itx-|x|}\mathrm{d}x$$
$$=\frac{1}{2}\int_0^\infty e^{(-it+1)-x}\mathrm{d}x+\frac{1}{2}\int_{-\infty}^0e^{(it+1)x}\mathrm{d}x$$
$$=\frac{1}{2}e^{(-it+1)}\int_0^\infty e^{-x}\mathrm{d}x+\frac{1}{2}e^{(it+1)}\int_{-\infty}^0e^{x}\mathrm{d}x$$
$$\frac{1}{2}e^{-it+1}+\frac{1}{2}e^{(it+1)}$$
$$\frac{1}{2}e(e^{-it}+e^{it})$$
But this doesn't look like the characteristic function on Wikipedia.

First I will derive what we will need later using Euler's formula.
\begin{align*}
  \int \cos(ux) e^{-x} dx + i \int \sin(ux) e^{-x} dx&= \int (\cos(ux) + i \cdot \sin(ux)) e^{-x} dx \\
  &= \int e^{iux} \cdot e^{-x} dx \\
  &= \int e^{(iu-1)x} dx \\
  &= \frac{1}{iu-1} e^{(iu-1)x} \\
  &= \frac{1}{iu-1} \frac{(iu+1)}{(iu+1)} e^{(iu-1)x} \\
  &= -\frac{e^{-x}}{u^2+1} (iu+1) \cdot e^{iux} \\
  &= -\frac{e^{-x}}{u^2+1} (iu+1) \cdot (\cos(ux) + i \sin(ux)) \\
  &= -\frac{e^{-x}}{u^2+1} (iu\cos(ux) - u\sin(ux) + \cos(ux) + i \sin(ux)) \\
  &= -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)) + i (-\frac{e^{-x}}{u^2+1} (u\cos(ux) + \sin(ux))).
\end{align*}
Taking the real term,
$\int \cos(ux) e^{-x} dx = -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)). \qquad (1)$
Now, the characteristic function is
\begin{align*}
  \varphi_{X} (u) &= E[e^{iuX}] \\
              &= \int_{-\infty}^{\infty} e^{iux} \cdot \frac{1}{2} e^{-|x|} dx \\
              &= \frac{1}{2} \int_{-\infty}^{0} e^{iux} \cdot e^{-(-x)} dx + \frac{1}{2} \int_{0}^{\infty} e^{iux} \cdot e^{-x} dx \\
              &= \frac{1}{2} \int_{-\infty}^{0} (\cos(ux) + i \sin(ux)) \cdot e^{x} dx + \frac{1}{2} \int_{0}^{\infty} (\cos(ux) + i \sin(ux)) \cdot e^{-x} dx \\
              &= \frac{1}{2} \int_{0}^{\infty} (\cos(ux) - i \sin(ux)) \cdot e^{-x} dx + \frac{1}{2} \int_{0}^{\infty} (\cos(ux) + i \sin(ux)) \cdot e^{-x} dx \\
              &= \int_{0}^{\infty} \cos(ux) \cdot e^{-x} dx \\
              &= -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)) \Big|_{0}^{\infty} \qquad (1) \\
              &= 0 - (- \frac{1}{u^2+1} (1 - 0)) \\
              &= \frac{1}{u^2+1}.
\end{align*}
