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 $C$?



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