I'm looking for a "polite" way to calculate this integral using Laplace transform:
$$ \int_0^{+\infty} \frac{e^{-ax} – e^{-bx} }{x} dx. $$
Now the impolite way is to invoke a famous theorem from basic differential equations:
$$ \mathcal{L}( f(x)/x) = \int_s^\infty F(t) dt,$$
Where $ F$ is the LT of $f$. Setting $f(x)/x= 1/x$ in this theorem (and recalling the definition of LT) gives
$$ \int_0^\infty \frac{e^{-ax}}{x} = \infty – \ln(a).$$
Similarly
$$ \int_0^\infty \frac{e^{-bx}}{x} = \infty – \ln(b).$$
Putting the results together and cancelling infinity gives the final result $\ln \frac{b}{a}$ which I know for a fact is the true value of the integral. But I also know too well that cancelling infinity is not allowed. So what is the trick?
Best Answer
$$g(x) = \frac{e^{-ax} - e^{-bx} }{x} = \frac{(1-e^{-bx}) - (1-e^{-ax}) }{x}= \frac{f_b(x)}{x} - \frac{f_a(x)}{x}$$ where
$f_c(x) = (1- e^{-cx})$ for $c \gt 0$. Then denoting the step function by $u$ we have $\mathcal L(f_c \cdot u)(s) = \frac{c}{s(s+c)}$ and
$$ \begin{aligned}\mathcal{L}( g)(s) &= \int_s^\infty F(t) dt = \int_s^\infty \left(\frac{b}{t(t+b)}-\frac{a}{t(t+a)}\right) \ dt\\ &= \int_s^\infty \left(\frac{1}{t+a}-\frac{1}{t+b}\right) \ dt\\ &= \ln\left(\frac{s+b}{s+a}\right) \end{aligned}$$
The desired equality is finally obtained by plugging in $s=0$ in previous equation.