A Clever Integration Trick

This trick is from Integration for Engineers and Scientists by William Squire via The Handbook of Integration by Daniel Zwillinger.

Noting that

\frac{1}{x} = \int\limits_{0}^{\infty} \mathrm{e}^{-xt} \mathrm{d} t

we can replace $$\frac{1}{x}$$ in an integrand with its integral expression and reverse the order of integration to simplify evaluation of the integrals. Of course the integral must converge uniformly to allow us to reverse the order of integration and we must have $$t>0$$.

Let us use this trick to evaluate

\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x} \mathrm{d} x

for $$a,b > 0$$.
Using our trick yields
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x} \mathrm{d} x & = \int\limits_{0}^{\infty}\int\limits_{0}^{\infty} \mathrm{e}^{-xt} (\mathrm{e}^{-ax}-\mathrm{e}^{-bx}) \mathrm{d} t \mathrm{d} x \\
& = \int\limits_{0}^{\infty}\int\limits_{0}^{\infty} \mathrm{e}^{-(a+t)x}-\mathrm{e}^{-(b+t)x} \mathrm{d} x \mathrm{d} t \\
& = -\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-(a+t)x}}{a+t} – \frac{\mathrm{e}^{-(b+t)x}}{b+t} |_{0}^{\infty} \mathrm{d} t \\
& = \int\limits_{0}^{\infty} \frac{1}{a+t} – \frac{1}{b+t} \mathrm{d} t \\
& = \lim_{R \to \infty} \mathrm{ln}\frac{a+t}{b+t} |_{0}^{R} \\
& = \mathrm{ln}\left(\frac{b}{a}\right)
\end{align}

The conventional way to handle this integral is to recognize that it is the Frullani integral

\int\limits_{0}^{\infty} \frac{f(ax) – f(bx)}{x} \mathrm{d} x = [f(\infty) – f(0)]\mathrm{ln}\left(\frac{a}{b}\right)

A simple substitution yields our result.

For the Frullani integral, we must have the existence of $$f(\infty)$$ and $$f(0)$$ using the appropriate limits. For other conditions on the integral as well as proofs, see

1. On Cauchy-Frullani Integrals by A. M. Ostrowski. Use DOI 10.1007/BF02568143 with Sci-Hub to access the paper.
2. On the Theorem of Frullani by Juan Arias-De-Reyna. Use DOI 10.2307/2048376 with Sci-Hub to access the paper.