TITLE: Math Subject GRE 1268 Question 55
QUESTION [16 upvotes]: If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$.  
A: $0$
B: $1$
C: $a-b$
D: $(a-b)\log 2$
E: $\frac{a-b}{ab}\log 2$
I really don't see how to start this one, I'm not so great with integrals.

REPLY [3 votes]: Another approach. $$\int_{0}^{\infty}\frac{e^{-bx}}{1+e^{-bx}}dx=\int_{0}^{\infty}\sum_{k\geq1}\left(-1\right)^{k+1}e^{-kbx}dx=\sum_{k\geq1}\left(-1\right)^{k+1}\int_{0}^{\infty}e^{-kbx}dx
 $$ $$=\frac{1}{b}\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}=\frac{1}{b}\log\left(2\right)
 $$ and in similar way we can calculate $\int_{0}^{\infty}\frac{e^{-ax}}{1+e^{-ax}}dx
 $. So 

$$
\color{blue}{\int_{0}^{\infty}\frac{e^{bx}-e^{ax}}{\left(1+e^{bx}\right)\left(1+e^{ax}\right)}dx=\left(\frac{1}{b}-\frac{1}{a}\right)\log\left(2\right).}$$