working on a problem to evaluate
$\int_0^\infty \frac {e^{-x} - e^{-ax}} {x} dx$
the instructions say to first evaluate $\int_1^a e^{-xy} dy$
which comes out to the integrand of the original improper integral.
using this it seems we can rewrite the improper integral as the double integral as follows
$\int{_0^\infty}\int{_1^a} e^{-xy}dydx$
However where I am stuck is changing the order of integration on this double integral in order to evaluate it.
Let's start$$\begin{align*}\int\limits_0^{\infty}\mathrm dx\,\frac {e^{-x}-e^{-ax}}x & =\int\limits_0^{\infty}\mathrm dx\,\int\limits_1^a\mathrm dy\, e^{-xy}\\ & =\int\limits_1^a\mathrm dy\,\int\limits_0^{\infty}\mathrm dx\, e^{-xy}\end{align*}$$Letting $z=xy$ and evaluating the integral gives that$$\int\limits_0^{\infty}\mathrm dx\, e^{-xy}=\frac 1y$$Hence$$\int\limits_0^{\infty}\mathrm dx\,\frac {e^{-x}-e^{-ax}}x\color{blue}{=\log a}$$
