Evaluate the integral: $\int_0^\infty \frac{e^{-ay}-e^{-by}}{y}\mathrm{d}y$ (trying to understand the substitution for Feynman's method for this... I've seen a few examples of exactly this problem, but I can't seem to get the mechanics of it:
$$\int_0^\infty \frac{e^{-ay}-e^{-by}}{y}\mathrm{d}y$$

*

*splits to $\int \frac{e^{-ay}}{y}\mathrm{d}y - \int \frac{e^{-by}}{y}\mathrm{d}y$ ?

*add some function?

 A: Note that $$\frac{e^{-ay} - e^{-by}}{y} = \int_{a}^{b} \exp(-yz) \, dz$$ so we have $$\int_{0}^{\infty} \int_{a}^{b} \exp(-yz) \, dz \, dy \overset{\text{Fubini}}{=} \int_{a}^{b} \int_{0}^{\infty} \exp(-yz) \, dy \, dz $$ $$= \int_{a}^{b} \frac{1}{y} \, dy = \ln(b/a)$$ The use of Fubini's theorem is justified since the function of interest is nonnegative everywhere (and hence Tonelli's theorem applies).
A: Let $I$ be the original integral. To use Feynman's Method, we can use the Laplace Transformation
$$F\left(s\right)\ =\ \int_{0}^{\infty}\frac{e^{-ay}-e^{-by}}{y}e^{-sy}dy,$$
where $s+a,s+b > 0$. Taking $\dfrac{d}{ds}$ on both sides, we get
$$F'(s) = \int_0^{\infty}\frac{\partial}{\partial s}\frac{e^{-ay}-e^{-by}}{y}e^{-sy}dy = -\int_{0}^{\infty}\left(e^{-ay}-e^{-by}\right)e^{-sy}dy = \frac{a-b}{s^{2}+\left(a+b\right)s+ab}.$$
Integrating from $0$ to $\infty$ on both sides with respect to $s$, we apply the Fundamental Theorem of Calculus to get
$$\int_{0}^{\infty}F'\left(s\right)ds\ =\ F\left(\infty\right)-F\left(0\right)\ =\ 0\ -\ I\ =-I.$$
We can also do some grunt work and get
$$\int_{0}^{\infty}F'\left(s\right)ds\ =\ \int_{0}^{\infty}\frac{a-b}{s^{2}+\left(a+b\right)s+ab}ds\ =\ \ln\left(a\right)-\ln\left(b\right).$$
By the transitive property, we get
$$-I = \ln\left(a\right)-\ln\left(b\right).$$
Therefore, the integral $I$ is
$$\int_{0}^{\infty}\frac{e^{-ay}-e^{-by}}{y}dy = \ln\left(b\right) - \ln\left(a\right).$$
