What is the characterstic function?

Question

Good evening,

what is the characteristic function of a distribution with density function $$f(x) = \frac{1}{2\lambda}e^{-\frac{|x|}{\lambda}},$$ while $x\in\mathbb{R}$ with parameter $\lambda > 0$?

Answer 1

As yet I've got:

$$\frac{1}{2\lambda}\exp\left(-\frac{|x|}{\lambda}\right) = \begin{cases} \frac{1}{2\lambda}\exp\left(-\frac{x}{\lambda}\right)&, x\geq 0 \\ \frac{1}{2\lambda}\exp\left(\frac{x}{\lambda}\right)&,x<0 \end{cases}$$

Now, I define my characteristic function

\begin{align*} \varphi_X(t)&=\mathbb{E}(e^{itX}) \\ &=\int_{-\infty}^{+\infty} e^{itx}\frac{1}{2\lambda}\exp\left(-\frac{|x|}{\lambda}\right) dx \\ &=\frac{1}{2\lambda}\int_{-\infty}^{+\infty} e^{itx}\exp\left(-\frac{|x|}{\lambda}\right) dx \\ &=\frac{1}{2\lambda}\left(\int_{-\infty}^0 e^{itx}\exp\left(\frac{x}{\lambda}\right) + \int_{0}^{+\infty} e^{itx}\exp\left(-\frac{x}{\lambda}\right)\right). \end{align*} I know that $e^{itx} = \cos(tx)+i\sin(tx)$ and I look the first integral:

\begin{align*} \int_{-\infty}^0 e^{itx}\exp\left(\frac{x}{\lambda}\right) &= \int_{-\infty}^0(\cos(tx)+i\sin(tx))\exp(\frac{x}{\lambda}dx \\ &= \int_{-\infty}^0 \cos(tx)\exp(\frac{x}{\lambda})dx + i\int_{-\infty}^0 \sin(tx)\exp(\frac{x}{\lambda})dx \end{align*} After the calculation I have: $$\varphi_X(t) = \frac{(1-i)\lambda t}{\lambda^2t^2+1}$$ Is it right?