Does this integral converge and can it be solved or approximated analytically?

Question

I have the following integral:

$Q=\int_{t=T_2}^{\infty}\left(\frac{Ke^{-A(T+t)}\left(1-\frac{1}{(CG-1)(-1+(-1+e^{At})(F-1)G}\right)}{1+(-1+e^{-A(T+t)})F}\right)dt$, where $0<C<1$, $0<F \leq 1$, $0<G \leq 1$, and $A,K,T, T_2>0$.

Is there a way to show whether this integral converges? If so, can this integral be solved analytically? Perhaps using an approximation of some sort?

Thus far, I have found that the limit of $Q$ as $t \rightarrow 0$ is $\frac{-CGK}{(e^{AT}(F-1)-F)(CG-1)}$ and as $t \rightarrow \infty$ is zero.

I have tried simply solving the integral with WolframOnline, but the computation exceeds the time limit for my plan.

Answer 1

Using Mathematica:

(Integrate[k Exp[-a (t + tt)] (1 - 1/((c g - 1) (-1 + (-1 + Exp[a t]) (f - 1) g)))/
  (1 + f (-1 + Exp[-a (t + tt)])), {tt, t2, \[Infinity]},
  Assumptions -> {0 < c < 1, 0 < f <= 1, 0 < g <= 1, a > 0, t > 0, t2 > 0, k > 0}]) /. 
{a -> A, c -> C, f -> F, g -> G, t2 -> T2, t -> T}

$$\frac{k \left(1-\frac{1}{(C G-1) \left((F-1) G \left(e^{A T}-1\right)-1\right)}\right) \left(\log \left(e^{A (T+\text{T2})}-\frac{F}{F-1}\right)-A (T+\text{T2})\right)}{A F}$$

The last replacement part {a -> A, c -> C, f -> F, g -> G, t2 -> T2, t -> T} is because one should avoid using uppercase letters to start variable names in Mathematica and that use of that replacement rule was only to match the original variable names without having to edit the LaTeX code by hand.