I am trying to find the integral of the following expression, with respect to $t$ from $0$ to infinity:
$$\frac{e^{-a g t} \left(b-(g-1) \left(\text{H0} \left(e^{a t (-b+g- 1)}-1\right)+1\right)\right) \left(b \text{Wi} \left(e^{a (b+1) t}-1\right)+(b+1) \text{Wh}\right)}{(b-g+1) \left(b e^{a (b+1) t}+1\right)}$$
$a$, $b$, and $g$ are all real and positive. I don't have much training in calculus. So I have been using mathematica to solve this. I get the following solution
It is quite complicated... just wondering if there is a more elegant solution to the integral?
$\text{ConditionalExpression}\left[\frac{g (b+(g-1) (\text{H0}-1)) \, _2F_1\left(1,\frac{g}{b+1}+1;\frac{g}{b+1}+2;-\frac{1}{b}\right) \Gamma \left(\frac{b+g+1}{b+1}\right) (b \text{Wh}-b \text{Wi}+\text{Wh})+b \Gamma \left(\frac{2 b+g+2}{b+1}\right) \left((b+1) \text{Wi} (b+(g-1) (\text{H0}-1)) \, _2F_1\left(1,\frac{g}{b+1};\frac{g}{b+1}+1;-\frac{1}{b}\right)+(g-1) g \text{H0} (b (b+1) \log (-b-1) (\text{Wh}-\text{Wi})-b (b+1) \log (-b) (\text{Wh}-\text{Wi})-b \text{Wh}+b \text{Wi}-\text{Wh})\right)}{a b (b+1) g (b-g+1) \Gamma \left(\frac{2 b+g+2}{b+1}\right)},\left(\Re\left((-b)^{\frac{1}{a b+a}}\right)\geq 1\lor \Re\left((-b)^{\frac{1}{a b+a}}\right)\leq 0\lor (-b)^{\frac{1}{a b+a}}\notin \mathbb{R}\right)\land \Re(a (b+1))>0\land \Re(a g)>0\land \Re(a (b+g+1))>0\right]$
I am not sure this will be much better that what you obtained $$\frac{(g-1) \text{H0} (b (b+1) (\log (-b-1)-\log (-b)) (\text{Wh}-\text{Wi})+(-b-1) \text{Wh}+b \text{Wi})+\left(-\frac{1}{b}\right)^{-\frac{g}{b+1}} (b+(g-1) (\text{H0}-1)) \left(\text{Wi} B_{-\frac{1}{b}}\left(\frac{g}{b+1},0\right)-B_{-\frac{1}{b}}\left(\frac{g}{b+1}+1 ,0\right) (b \text{Wh}-b \text{Wi}+\text{Wh})\right)}{a (b+1) (b-g+1)}$$