How to prove that the integral $\int_{0}^\infty \left[1-\frac{dT}{(Q-1)[(e^{T}-1)(F-1)G-1]}\right]$ does not converge? I have the following integral:
$$\int_{0}^\infty \left[1-\frac{dT}{(Q-1)[(e^{T}-1)(F-1)G-1]}\right],$$ where $0<Q<1$, $T \geq 0$, $0<G \leq 1$, and $0<F < 1$.
Mathematica says that this integral does not converge, so I am trying to understand why.
Thus far, I have found that the limit as $T \rightarrow 0$ has a finite value of $1+\dfrac{1}{Q-1}$ and the limit as $T \rightarrow \infty$ is one. And the denominator is always positive under the given parameter constraints, which I interpret to mean that the integrand is a continuous function.
Any suggestions as to how I can show that this integral does not converge would be greatly appreciated!
 A: You wrote the integral in a very bad way. First of all, the measure $\text{d}T$ cannot be written as you did. It must be outside the whole square bracket, otherwise the integral is ill-defined because the first term, the $1$ alone, has to be integrated with respect to what? And of course Mathematica tells you the integral does not converge, for
$$\int_0^{+\infty} 1 = +\infty$$
(no matter if it's $\text{d}T$ or $\text{d}$-else).
Secondly, we can write your integral in a rather more pleasant way, noticing that:
$$(Q-1)[(e^T - 1)(F-1)G - 1] = (Q-1)[(e^T-1)\beta - 1]$$
Where $\beta = (F-1)G$. Continuing to arrange and naming new variables we get
$$Ae^T - B$$
where $A = \alpha\beta$, $B = \alpha\beta - \alpha$ and again $\alpha = Q-1$ and $\beta = (F-1)G$
So the integral you have can we written as
$$J = \int_0^{+\infty} \left(1 - \frac{1}{Ae^T - B}\right)\ \text{d}T$$
Now you can take Hafez answer as good, observing that indeed
$$\text{as}\ T\to +\infty, \quad J \sim 1$$
$$\text{as}\ T\to 0, \quad J \sim 1 + \frac{1}{B}$$
Ans so on.
Consider that the problem of its convergence does not exactly lies in the lower bound of the integral. For example, the same integral from $0$ to $1$ has a closed form, that is
$$\int_0^1 \left(1 - \frac{1}{Ae^T - B}\right)\ \text{d}T = \frac{\log (B (B-A))-\log (B (B-e A))+1}{B}+1$$
With the many conditions of existence upon $A$ and $B$
Where as if you extend the upper limit, you can see a great dependence on the upper bound. Say
$$\int_0^M \left(1 - \frac{1}{Ae^T - B}\right)\ \text{d}T = \frac{-\log \left(B \left(B-A e^M\right)\right)+\log (B (B-A))+\log \left(e^M\right)}{B}+M$$
(which holds iff $e^M \geq 0$). You can see here that for $M \to +\infty$ the integral diverges, but there are no problems in $0$.
