Implication of Lévy-Khintchine theorem/representation I have trouble understanding the use/implication of the Lévy-Khintchine theorem. One possible way to state it is the following:
The characteristic function $\varphi$ is infinitely divisible if and only if $\varphi = \exp{\psi}$ and
$$\psi(\lambda) = i \lambda a - \frac{\sigma^2 \lambda^2}{2} + \int_{\mathbb R} \left(\exp(i \lambda x) - 1 - \frac{i\lambda x}{1+x ^2}\right)\frac{1+x^2}{x^2} \nu(dx)$$
where $\nu$ is a finite measure on $\mathbb R$ with $\nu\{0\}=0$,
$(a,\sigma^2,\nu)$ is the Lévy-Khintchine triplet.
Now, this formula looks a little intimidating. What would be its value/implications in words?
Somehow, one can use the theorem to decide the following question:
Is there a time-homogenous, stochastically continuous process $X(t)$ with independent increments and with $X(1)\sim Exp(1)$?
I unfortunately do not see how the theorem might help answer this question...
 A: 
Let $\nu$ a measure on $\mathbb{R}$ such that $\nu(\{0\})=0$ and $a,\sigma^2 \in \mathbb{R}$. Define $\psi$ by $$\psi(\lambda) = i \lambda a - \frac{\sigma^2 \lambda^2}{2} + \int_{\mathbb R} \left(\exp(i \lambda x) - 1 - \frac{i\lambda x}{1+x ^2}\right)\frac{1+x^2}{x^2} \nu(dx) \tag{1}$$
Then there exists a time-homogenous stochastically continuous process $(X_t)_t$ with independent increments such that $$\mathbb{E}e^{\imath \, \lambda \cdot X_t} = \exp(t \cdot \psi(\lambda))$$ in particular $\mathbb{E}e^{\imath \, \lambda \cdot X_1} = \exp(\psi(\lambda))$.

Thus, it suffices to find a finite measure $\nu$ and $b, \sigma^2 \in \mathbb{R}$ such that $$\exp(\psi(\lambda)) = \mathbb{E}e^{\imath \, \lambda \cdot X_1} = \mathbb{E}e^{\imath \, \lambda \cdot \text{Exp}(1)} = \frac{1}{1-\imath \, \lambda} = \exp(-\log(1-\imath \lambda))$$ where $\log$ denotes the principal value (so the imaginary part is in $(-\pi,\pi]$).
To find a suitable measure $\nu$ we start rewriting the expression $-\log(1-\imath \, t)$: $$\begin{align} -\log(1-\imath \, \lambda) &= \imath \int_0^{\lambda} \frac{1}{1-\imath \, t} \, dt = \imath \int_0^{\lambda} \int_0^{\infty} e^{-x \cdot (1-\imath \, t)} \, dx \, dt \\ &= \imath \, \int_0^{\infty} \int_0^{\lambda} e^{-x \cdot (1-\imath \, t)} \, dt \, dx = \int_0^{\infty} \frac{e^{\imath \, \lambda \cdot x}-1}{x} \cdot e^{-x} \, dx \tag{2} \end{align}$$
This already looks a bit similar to $(1)$. Indeed, $$\begin{align} - \log(1-\imath \, t) &\stackrel{(2)}{=} \int_0^{\infty} \frac{e^{\imath \, \lambda \cdot x}-1}{x} \cdot e^{-x} \, dx = \int_0^{\infty} \left(e^{\imath \, \lambda \cdot x}-1 - \frac{\imath \, \lambda \cdot x}{1+x^2} \right) \cdot \frac{e^{-x}}{x} \, dx + \imath \, \lambda \cdot a \end{align}$$ where $$a := \int_0^{\infty} \frac{e^{-x}}{1+x^2} \, dx$$
So we arrive at $- \log(1-\imath \, \lambda) = \psi(\lambda)$ where $$\begin{align} a &= \int_0^{\infty} \frac{e^{-x}}{1+x^2} \, dx \\ \sigma^2 &= 0 \\ \nu(dx) &:= \frac{x}{1+x^2} \cdot e^{-x} \cdot 1_{(0,\infty)}(x) \, dx \end{align}$$ Obviously, $\nu$ is a finite measure, $\nu(\{0\})=0$. By applying the theorem, we conclude that there exists a time-homogenous stochastically continuous process $(X_t)_t$ with independent increments such that $X_1 \sim \text{Exp}(1)$.
