In the following Eq. 1, the integral diverges. However, it is part of my equation for the Expectation of rate and it should not diverge. I will really appreciate any help as I am stuck on this for over 4 days now.
$$\begin{align} E = \int_r^\infty\dfrac{\mathrm{e}^{-x}}{1+ \mathrm{e}^{b-cx}}dx \tag{1} \label{1} \end{align} $$
In the following, I found I can solve it using the My Take below, but even that diverges.
My Take:
--> following the properly $(1+z)^{-1} = \sum_{j=1}^{\infty} (-1)^{j-1}(z)^{j-1}$
$$E = \int_r^\infty\mathrm{e}^{-x}\sum_{j=1}^{\infty} (-1)^{j-1}(e^{b-cx})^{j-1}dx$$
$$E = \sum_{j=1}^{\infty} (-1)^{j-1} e^{b(j-1)}\int_r^\infty\mathrm{e}^{-x}.e^{-cx(j-1)} dx$$
$$E = \sum_{j=1}^{\infty} (-1)^{j-1} e^{b(j-1)}\int_r^\infty\mathrm e^{-x-cx(j-1)}dx$$
$E$ diverge even now.
How can I solve this?
Provided that $b,c>0$ we have $$\frac{e^{-x}}{1+e^{b-c x}}=\frac{1}{e^x+e^{b+(1-c) x}}<\frac{1}{e^b\left(e^{x-b}+e^{x-cx}\right)}<\frac{1}{ke^{x}};\;k>0$$ Therefore the integral converges for the comparison test.