Calculating a certain definite integral

Question

How to calculate the following :

$$\int_{0}^{\infty}\frac{1}{1+e^{ax}}-\frac{1}{1+e^{bx}}\text{ }\mathrm{d}x$$

Answer 1

\begin{align} \tag{a} \int \frac{1}{1+\mathrm{e}^{ax}} dx &= \frac{1}{a} \int \frac{1}{u(u-1)} du \\ \tag{b} &= \frac{1}{a} \int \left(\frac{1}{u-1}-\frac{1}{u}\right) du \\ &= \frac{1}{a} \Big[\mathrm{ln}(u-1) - \mathrm{ln}(u) \Big] \\ &= \frac{1}{a} \Big[ax - \mathrm{ln}(1+\mathrm{e}^{ax}) \Big] \end{align}

Applying the limits of integration we obtain $$\int\limits_{0}^{\infty} \frac{1}{1+\mathrm{e}^{ax}} dx = \frac{1}{a} \mathrm{ln}(2)$$

Thus we have \begin{equation} \int\limits_{0}^{\infty} \frac{1}{1+\mathrm{e}^{ax}} - \frac{1}{1+\mathrm{e}^{bx}} = \left(\frac{1}{a} - \frac{1}{b}\right) \mathrm{ln}(2) \end{equation}

a. $u=1+\mathrm{e}^{ax}$

b. partial fractions

Answer 2

Suggestion: split the integral into a difference of two integrals. Multiply through the denominator and numerator of the first integrand with $ e^{-ax}$, and do the equivalent for the second integral. At that point, think substitution. Is that enough?

Answer 3

The integral has a closed form for $a>0$: $$\int_0^\infty \frac{1}{1+e^{a x}}dx=\frac{\ln 2}{a}$$