Find the characteristic function $\phi_X(t)$ of an absolute continious r.v. $X$ with density:
$$f_X(x) = \frac{a}{2}e^{-a|x|} \qquad (a>0; x\in \mathbb{R})$$
Notation
I have some problems writing this down, how should I handle the complex number?
$$\begin{align} \phi_X(t) &= \operatorname{E}(e^{itX})\\ & = \int_{-\infty}^{+\infty} e^{itx}\frac{a}{2}e^{-a|x|}\operatorname{d}x\\ & = \frac{a}{2}\left( \int_{0}^{+\infty} e^{itx}e^{-ax}\operatorname{d}x+\int_{-\infty}^{0} e^{itx}e^{ax}\operatorname{d}x\right)\\ & = \frac{a}{2}\left( \int_{0}^{+\infty} e^{x(it-a)}\operatorname{d}x+\int_{-\infty}^{0} e^{x(it+a)}\operatorname{d}x\right)\\ & = \frac{a}{2}\left( \left[ \frac{e^{x(it-a)}}{it-a}\right]^{+\infty}_0+\left[ \frac{e^{x(it+a)}}{it+a}\right]^0_{-\infty}\right)\\ \end{align}$$
However this notation doesn't seem to make any sense... How should I write this down?
Since $a>0$ we have
$$\left| e^{r (\imath t-a)} \right| = e^{-r \cdot a} \to 0 \qquad \text{as} \, \, r \to \infty$$
for any $t \in \mathbb{R}$. Therefore,
$$\begin{align*} \left[ \frac{e^{x(\imath t-a)}}{\imath \, t-a} \right]_0^{\infty} &= \lim_{r \to \infty} \left[ \frac{e^{x(\imath t-a)}}{\imath \, t-a} \right]_0^{r} \\ &= \frac{1}{\imath \, t-a} \lim_{r \to \infty} (e^{r (\imath \, t-a)}-1) \\ &= - \frac{1}{\imath \, t-a}. \end{align*}$$
A similar argumentation applies to the second term.