I am trying to evaluate the following integral
$$\int_0^\infty \frac{e^{ax}-e^{bx}}{\left(1+e^{ax}\right)\left(1+e^{bx}\right)}dx.$$
After multiplying the numerator and denominator of the integrand by $e^{-\frac{1}{2}(a+b)}$ and using partial fractions decomposition, I arrived at two other the equivalent expressions:
$$\int_0^\infty \frac{\sinh\left[\frac 12(a-b)x\right]}{\cosh\left(\frac 12ax\right)\cosh\left(\frac12bx\right)}dx = \int_0^\infty \frac{\sinh\left(\frac 12ax\right)}{\cosh\left(\frac 12ax\right)} – \frac{\sinh\left(\frac 12bx\right)}{\cosh\left(\frac12bx\right)}dx.$$
The expression with the difference of hyperbolic tangents looks promising because the antiderivative of $\tanh ax$ is $a^{-1}\log(\cosh ax)$. Is this on the right track?
Best Answer
The integrand can be rewritten as
$$\int_0^\infty \frac{1+e^{ax}-1-e^{bx}}{(1+e^{ax})(1+e^{bx})}\;dx = \int_0^\infty \frac{1}{1+e^{bx}}-\frac{1}{1+e^{ax}}\:dx$$
$$= -\frac{1}{b}\log(1+e^{-bx})+\frac{1}{a}\log(1+e^{-ax})\Biggr|_0^\infty = \left(\frac{1}{b}-\frac{1}{a}\right)\log 2$$