Let $a$ and $b$ be positive numbers. We want to calculate the value of the improper integral

(1)

The value may be found e.g. by utilising the derivative of the integral $$I(y) \;:=\, \int_0^\infty e^{-xy}\!\cdot\!\frac{e^{-ax}-e^{-bx}}{x}\,dx$$ which can be formed by differentiating under the integral sign:

There is another method via Laplace transforms. By the table of Laplace transforms, we have $$\mathcal{L}\{e^{-at}-e^{-bt}\} \;=\; \frac{1}{s\!+\!a}-\frac{1}{s\!+\!b}$$ and therefore $$\mathcal{L}\{\frac{e^{-at}-e^{-bt}}{t}\} \;=\; \int_s^\infty\left(\frac{1}{u\!+\!a}-\frac{1}{u\!+\!b}\right)\,du \;=\; \sijoitus{u=s}{\quad\infty}\ln\frac{u\!+\!a}{u\!+\!b} \;=\; \ln\frac{s\!+\!b}{s\!+\!a},$$ i.e. $$\int_0^\infty e^{-st}\!\cdot\!\frac{e^{-at}-e^{-bt}}{t}\,dt \;=\; \ln\frac{s\!+\!b}{s\!+\!a}.$$ Letting $s \to 0+$ , this yields the equation $$\int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}\,dx \;=\; \ln\frac{b}{a}.$$