Laplace distribution has PDF: $f_X(x) = \frac{1}{2} e^{-|x|}$. I'm trying to calculate its characteristic function:
$$h_X(t) = E(e^{itX}) = \int \limits_\mathbb{R} e^{itx}f_X(x) \ dP(x)$$
Since the density function is symmetric, I wrote the above as doubled integral from zero to infinity:
$$ = 2 \int_0^\infty e^{itx} \cdot \frac 1 2 \cdot e^{-x} \ dP(x)$$
which led me to the improper integral:
$$\int_0^\infty e^{x(it-1)} \ dx = \frac 1 {it - 1} \left ( \lim_{x \rightarrow \infty} e^{-x+ixt} - 1 \right )$$
However, I do not know how to calculate the last limit. Could you give me any hint?
The density is symmetric, but the function you're integrating is not symmetric, since the factor $e^{itx}$ is not symmetric. However, that can be fixed by treating the two halves of the line separately. Both are dealt with in the same way except for small (but important) details. $$ |e^{-x+xit}| = |e^{-x}| |e^{xit}| = e^{-x} \cdot 1 \to 0 \text{ as } x\to\infty. $$ If the absolute value of a function of $x$ goes to $0$ then that function of $x$ goes to $0$.
PS: The notation $$\int_\mathbb{R} e^{itx}f_X(x) \ dP(x)$$ is somewhat odd. If $P$ is a probabilty measure then one can write $$ \operatorname{E}(e^{itX}) = \int_\mathbb{R} e^{itx}\,dP(x). $$ But in this case the probability measure is $dP(x) = f_X(x)\,dx,$ so you have $$ \int_\mathbb{R} e^{itx} \, dP(x) = \int_\mathbb{R} e^{itx} f_X(x)\,dx $$ with no $\text{“}dP(x)\text{''}$ in that last integral.