Try multivariable calculus

$\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\,dx \; = \; \int_0^\infty \int_a^b e^{-xy}\,dy \,dx \; = \; \int_a^b \int_0^\infty e^{-xy}\,dx\,dy \; = \; \int_a^b y^{-1}\,dy \; = \; \ln(\tfrac{b}{a}) = \boxed{1}$ A standard combination of Tonelli's and Fubini's Theorems tells us that the reversal of order of iterated integration is valid.

Also known as a Frullani Integral ........

I personally used double feynman differentiation under the integral sign

This is how I did it, although I am not so sure if it is rigorous enough...

$F(a,b)=\displaystyle \int_{0}^{\infty} \dfrac{e^{-ax}-e^{-bx}}{x} \, dx \\ \dfrac{\partial F}{\partial a}=\displaystyle \int_{0}^{\infty} \dfrac{\partial }{\partial a} \left( \dfrac{e^{-ax}-e^{-bx}}{x} \right) \, dx =\displaystyle \int_{0}^{\infty} -e^{-ax} \, dx=\left[ \dfrac{e^{-ax}}{a} \right]_0^{\infty}= -\dfrac{1}{a} \\ \dfrac{\partial F}{\partial b}=\displaystyle \int_{0}^{\infty} \dfrac{\partial }{\partial b} \left( \dfrac{e^{-ax}-e^{-bx}}{x} \right) \, dx =\displaystyle \int_{0}^{\infty} e^{-bx} \, dx=\left[ \dfrac{-e^{-bx}}{b} \right]_0^{\infty}= \dfrac{1}{b} \\$

Integrating, $F(a,b)=\ln b - \ln a=\ln \frac{b}{a}=\boxed{1}$

This reminds me of Frullani's integral, henceforth the result.

Assuming that $(x|a)\in \mathbb{C}\land \Re(a)>0,\int \frac{e^{-a x}-e^{e (-a) x}}{x} \, dx \Longrightarrow ExpIntegralEi[-a x] - ExpIntegralEi[-a E x]$ .

$(x|a)\in \mathbb{C}\land \Re(a)>0,\underset{x\to \infty }{\text{lim}}(\text{Ei}(-a\ x)-\text{Ei}(-a\ e\ x)) \Longrightarrow 0$

$(x|a)\in \mathbb{C}\land \Re(a)>0,\underset{x\to 0}{\text{lim}}(\text{Ei}(-a\ x)-\text{Ei}(-a\ e\ x)) \Longrightarrow -1$

$0-(-1) \Longrightarrow 1$