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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?

K.defaoite
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User32563
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1 Answers1

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$$g(x) = \frac{e^{-ax} - e^{-bx} }{x} = \frac{(1-e^{-bx}) - (1-e^{-ax}) }{x}= \frac{f_b(x)}{x} - \frac{f_a(x)}{x}$$ 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.

mathcounterexamples.net
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