Show that $\int_0^\infty \frac{e^{ax}-e^{bx}}{\left(1+e^{ax}\right)\left(1+e^{bx}\right)}dx = \frac{a-b}{ab}\log 2$

Question

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?

Answer 1

The integrand can be rewritten as

$$\int_0^\infty \frac{1+e^{ax}-1-e^{bx}}{(1+e^{ax})(1+e^{bx})}\;dx = \int_0^\infty \frac{1}{1+e^{bx}}-\frac{1}{1+e^{ax}}\:dx$$

$$= -\frac{1}{b}\log(1+e^{-bx})+\frac{1}{a}\log(1+e^{-ax})\Biggr|_0^\infty = \left(\frac{1}{b}-\frac{1}{a}\right)\log 2$$

Answer 2

The original form you gave can immediately be written by partial fractions as $$ I = \int_0^\infty \Big[\frac{1}{1+e^{bx}}-\frac{1}{1+e^{ax}}\Big]dx. $$ One can use u-substitution to obtain $$ I = \frac{1}{a}\log\big|1+e^{a x} \big|-\frac{1}{b}\log\big|1+e^{a x} \big| \Bigg |_{x=0}^\infty.$$ Evaluating the limits provides $$ I = \Big(\frac{1}{b}-\frac{1}{a}\Big)\log(2)$$ as quoted.