# Inverting a Laplace transform (for a Lévy process)

Let $$\psi(\theta) = c\theta + \frac{\sigma^{2}}{2}\theta^{2} - \frac{\lambda\theta}{\alpha + \theta}.$$
For those who are wondering where this function comes from, $$\psi$$ is the Laplace exponent for a particular Lévy Process (Brownian motion with drift $$c$$ and volatility $$\sigma$$ added with a compound Poisson process of intensity $$\lambda$$ with exponential negative jumps of parameter $$\alpha$$. But it should not matter for my problematic.
Here: $$\sigma, \lambda, \alpha > 0$$, and $$c \ne 0$$.

Let $$q \geq 0$$. I want to show that the inverse Laplace transform of the function $$({\psi(\theta) - q})^{-1}$$ is given by:

$$W^{(q)}(x) = \frac{e^{\theta_1x}}{\psi'(\theta_{1})} + \frac{e^{\theta_2x}}{\psi'(\theta_{2})} + \frac{e^{\theta_3x}}{\psi'(\theta_{3})}$$,
where $$\theta_1 > 0 > -\theta_2 > -\theta_3$$ are the three solutions of the equation $$\psi(\theta) = q$$.
Note: It is easy to check that these $$\theta_i$$ exists.

Here's what I did so far:
By considering the partial fraction decomposition of $$({\psi(\theta) - q})^{-1}$$, we get:

$$({\psi(\theta) - q})^{-1} = \frac{A}{\theta - \theta_1} + \frac{B}{\theta + \theta_2} + \frac{C}{\theta + \theta_3}$$, A,B,C real numbers.
It is easy to apply the inverse Laplace exponent for this expression by using linearity of $$L$$ (Laplace operator).

If we have $$A = 1/\psi'(\theta_1)$$,
$$B = 1/\psi'(\theta_2)$$, and
$$C = 1/\psi'(\theta_3)$$, then the solution follows.

However, it seems a very nasty calculus exercise to verify that these are the $$A,B$$, and $$C$$ fitting to get $$({\psi(\theta) - q})^{-1}$$. (My friend and I tried, and we can't).

Has someone an idea on how could we get these $$A,B$$, and $$C$$?

Best,
Félix