Evaluate $\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx$

Question

Given the integral $\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx$

How can I evaluate it? Thanks!

Answer 1

Note that

$$\frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1 + e^{bx})} = \frac{1}{1 + e^{bx}} - \frac{1}{1 + e^{ax}}$$

Now integrate each piece separately, using the substitution $u = 1 + e^{bx}$ or $u = 1 + e^{ax}$.