TITLE: How to solve $\int_0^\infty \left(\sqrt{1-\gamma(t)}e^{c\gamma(t)}-1\right)dt$, where $\gamma(t) = \frac{e^{-at}}{b+e^{-at}}$?
QUESTION [6 upvotes]: $$\int_0^\infty \left(\sqrt{1-\gamma(t)}e^{c\gamma(t)}-1\right)dt,$$
where
$$\gamma(t) = \frac{e^{-at}}{b+e^{-at}}.$$
I'll be glad if anyone can help me with this. I already tried some obvious substitutions, unsuccessfully of course. I rather have an exact result, however, if this can only be solved approximatively I would ideally have an expansion in $b$.

REPLY [2 votes]: Assume $b$ is small, I want the leading term.
Let $v=\exp(-at)/b$ to get 
$$ \frac 1a\int_0^{1/b}\left[\sqrt{\frac1{1+v}}\exp(\frac{cv}{1+v})-1\right] \frac{dv}v$$
The function is bounded at $v=0$ and the main term is $O(v^{-3/2})$ for large $v$.  So the dominant part of the integral is $ \int \frac{-1}v dv$ for large $v$ near $v=1/b$, and that gives $\log b$.
So I think the dominant term is $(\log b)/a$
The next term would be $f(c)/a$ where 
$$f(c)/a=\frac1a\int_0^1[g(c,v)-1]\frac{dv}v+\frac1a\int_1^\infty g(c,v)\frac{dv}v$$
and after that a correction term
$$\frac1a\int_{1/b}^\infty g(c,v)\frac{dv}v$$
which is roughly $2\sqrt{b}\exp(c)/a$