How to integrate $\int_0^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1+ e^{bx})}dx$ where $a,b > 0$. This 
$$\
\int_0^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1+ e^{bx})}dx \text{ where } a,b > 0.
$$
is a problem that showed up on a GRE practice test. I believe you're supposed to use complex contour integration, but I'm not sure which contour to use. I think it's the keyhole contour, but I wasn't able to get anything useful.
 A: You do not need complex integration, just Fubini's theorem:
$$\begin{eqnarray*}I=\int_{0}^{+\infty}\left(\frac{1}{1+e^{bx}}-\frac{1}{1+e^{ax}}\right)\,dx&=&-\int_{0}^{+\infty}\int_{a}^{b}\frac{ x\, e^{xy}}{(1+e^{xy})^2}\,dy\,dx\\&=&-\int_{a}^{b}\int_{0}^{+\infty}\frac{ x\, e^{xy}}{(1+e^{xy})^2}\,dx\,dy\\&=&-\int_{a}^{b}\frac{\log 2}{y^2}\,dy\\&=&\color{red}{\left(\frac{1}{b}-\frac{1}{a}\right)\log 2}.\end{eqnarray*} $$
As a matter of fact, we just need to compute:
$$ \int_{0}^{+\infty}\frac{dx}{1+e^x}=\int_{1}^{+\infty}\frac{dt}{t(1+t)}=\int_{0}^{1}\frac{du}{1+u}=\log 2$$
that is straightforward by setting $x=\log t$, then $t=\frac{1}{u}$.
A: A real pedestrian approach would be as follows:
Let us define $J(c):=\int_{0}^{\infty}\frac{1}{1+e^{cx}}=\int_{0}^{\infty}\frac{e^{-cx}}{1+e^{-cx}}$, then we can apply geometric series to obtain
$$
J(c)=\int_{0}^{\infty}\frac{1}{1+e^{cx}}=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\infty}e^{-(n+1)cx}=\frac1c\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}
$$
Employing the Taylor series of $\log$ we get
$$
J(c)=\frac{\log(2)}{c}
$$
Now observe that the original integral $I(a,b)$ is given by
$$
I(a,b)=J(a)-J(b)=\log(2)\left(\frac{1}{a}-\frac{1}{b}\right)
$$
in agreement with all other answers
Contour integration is tricky here, maybe i will come up with something tomorrow
A: \begin{align}
\int_0^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1+ e^{bx})}dx&=\int_{0}^{\infty}\left(\frac{1}{1+e^{bx}}-\frac{1}{1+e^{ax}}\right)dx
\\
&=\int_{0}^{\infty}\left(\frac{e^{-bx}}{1+e^{-bx}}-\frac{e^{-ax}}{1+e^{-ax}}\right)dx
\\
&=-\frac1{b}\int_{0}^{\infty}\frac{1}{1+e^{-bx}}d(e^{-bx})+\frac1{a}\int_{0}^{\infty}\frac{1}{1+e^{-ax}}d(e^{-ax})
\\
&=-\frac1{b}\int_{1}^{0}\frac{1}{1+t}dt+\frac1{a}\int_{1}^{0}\frac{1}{1+t}dt
\\
&=\left(\frac1{b}-\frac1{a}\right)\log{2}
\end{align}
