= Solution
The <Fubini theorem> for an integrable Borel-measurable function states that, if $\int_{\mathbb R^2}|f|<\infty$, its sections are integrable for almost every coordinate and
$$
\int_{\mathbb R^2}f\,d(x,y)
=\int_{\mathbb R}\left(\int_{\mathbb R}f(x,y)\,dx\right)dy
=\int_{\mathbb R}\left(\int_{\mathbb R}f(x,y)\,dy\right)dx.
$$
For a nonnegative measurable function, <Tonelli theorem> gives the same equalities with infinity allowed before integrability is established.
Here $f$ is the <continuous> exponential multiplied by the indicator of a <Borel set>, so it is Borel-measurable and nonnegative. Tonelli and integration in $x$ give
$$
\int_{\mathbb R^2}f=\int_a^b\int_0^\infty e^{-xy}\,dx\,dy
=\int_a^b\frac{dy}{y}=\log(b/a)<\infty.
$$
This proves integrability. We may now integrate in the reverse order; for $x>0$ the inner integral is $(e^{-ax}-e^{-bx})/x$. Hence the <Frullani integral> is
$$
\boxed{\int_0^\infty\frac{e^{-ax}-e^{-bx}}x\,dx=\log\frac ba}.
$$
The apparent singularity at zero is removable in the integrand's limiting value $b-a$.
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