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Inverse Laplace transform to get CDF

I have the following problem. If i can get some help, I would greatly appreciate it.

Suppose X is a birth death process (represents population size) that evolves by:

$X -> X+1 $ if a birth occurs with rate $\mu$

$X -> X-1 $ if a death occurs with rate $\theta$

Suppose $T_A$ is first passage time of a BD process from state A to state 0 and suppose $T_B$ is first passage time of another BD process from state B to state 0. They are both independent.

I need to find $P(T_A < T_B)$. That is, probability that a population of size A goes to 0 before population of size B.

By definition:

$T_A = T_{A,A-1} + T_{A-1,A-2} + ... + T_{1,0}$

where $T_{i,i-1}$ represents first passage time from state i, to state i-1.

I read some articles online that mentioned that if $T = T_A - T_B$, then the CDF defined by:

$G_T(t) := P(T <=t)$

is what i need. The paper suggested taking inverse laplace of a CDF to obtain the CDF and evaluate it at 0. It first suggested finding laplace transform of T, which is given by

$L[T] = E(e^{-ST}) = -\frac{L[T_A](s) L[T_B](-s)}{s}$

Then it suggested taking laplace of $G_T(t)$ i.e $L[G_T(t)]$

However, $L[G_T(t)] = \frac{L[T]}{s} = \frac{L[T_A](s) L[T_B](-s)}{s}$

Then the paper suggests taking the inverse of above evaluated at 0 to get $P(T_A < T_B)$.

Question:

  1. Given the Laplace transform of CDF, which is $L[G_T(t)]$, I want to use the inverse laplace to obtain $G_T(t)$ evaluated at 0. But, by definition, inverse laplace using algorithms in python are all one sided from $[0,\infty]$ My random variable T is given by difference of two first passage times, $T = T_A - T_B $. Won't this be negative?

  2. If 1) is true, then how can i shift my region of integration from (-$\infty$,t) in a computer? should i multiply $L[G_T(t)]$ by a complex exponential?

Not sure if i am on the right track here

Thanks