We need to evaluate the definite integral $$ \int_{0}^{+\infty} \frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} \, dx $$

1. Problem Description

We need to evaluate the definite integral

$$$$

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

$$$$

2. Solution Steps

Let

$$$$

I = \int_{0}^{+\infty} \frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} \, dx

$$$$

We can rewrite the integrand as follows:

$$$$

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

$$$$

We can rewrite the first term as

$$$$

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

$$$$

And rewrite the second term as

$$$$

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

$$$$

Thus,

$$$$

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

$$$$

So the integral becomes

$$$$

I = \int_{0}^{+\infty} \left( \frac{1}{1+e^{bx}} - \frac{1}{1+e^{ax}} \right) dx

$$$$

Now, we can split the integral into two:

$$$$

I = \int_{0}^{+\infty} \frac{1}{1+e^{bx}} \, dx - \int_{0}^{+\infty} \frac{1}{1+e^{ax}} \, dx

$$$$

Let $u = bx$, so $x = u/b$ and $dx = \frac{1}{b} du$. When $x=0$, $u=0$. When $x \to \infty$, $u \to \infty$.

$$$$

\int_{0}^{+\infty} \frac{1}{1+e^{bx}} \, dx = \int_{0}^{+\infty} \frac{1}{1+e^{u}} \frac{1}{b} \, du = \frac{1}{b} \int_{0}^{+\infty} \frac{1}{1+e^{u}} \, du

$$$$

Similarly, let $v = ax$, so $x = v/a$ and $dx = \frac{1}{a} dv$. When $x=0$, $v=0$. When $x \to \infty$, $v \to \infty$.

$$$$

\int_{0}^{+\infty} \frac{1}{1+e^{ax}} \, dx = \int_{0}^{+\infty} \frac{1}{1+e^{v}} \frac{1}{a} \, dv = \frac{1}{a} \int_{0}^{+\infty} \frac{1}{1+e^{v}} \, dv

$$$$

Therefore,

$$$$

I = \frac{1}{b} \int_{0}^{+\infty} \frac{1}{1+e^{u}} \, du - \frac{1}{a} \int_{0}^{+\infty} \frac{1}{1+e^{v}} \, dv = \left( \frac{1}{b} - \frac{1}{a} \right) \int_{0}^{+\infty} \frac{1}{1+e^{x}} \, dx

$$$$

Now we need to evaluate $\int_{0}^{+\infty} \frac{1}{1+e^{x}} \, dx$. We can rewrite this as

$$$$

\int_{0}^{+\infty} \frac{e^{-x}}{e^{-x}+1} \, dx = \left[-\ln(1+e^{-x})\right]_{0}^{+\infty} = -\ln(1+0) + \ln(1+1) = \ln 2

$$$$

Thus,

$$$$

I = \left(\frac{1}{b} - \frac{1}{a}\right) \ln 2 = \left(\frac{a-b}{ab}\right) \ln 2

$$$$

Then

$$$$

I = \int_0^\infty \frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} dx = \int_0^\infty \left( \frac{1}{1+e^{bx}} - \frac{1}{1+e^{ax}} \right) dx = \frac{\ln 2}{b} - \frac{\ln 2}{a} = \frac{(a-b) \ln 2}{ab}

$$$$

3. Final Answer

$\frac{(a-b)\ln 2}{ab}$