I am reading the following [book](https://rads.stackoverflow.com/amzn/click/0122165500) and there is a step in a derivation I don't quite follow. In equation 27 on page 409 it is claimed


$$
\int\_0^{\infty}\,\epsilon^{\alpha}(1+\epsilon/x)^{-1}e^{-\epsilon}d\epsilon=x^{\alpha+1}e^x\int\_x^{\infty}d\zeta\int\_0^{\infty}\sigma^{\alpha}e^{-(1+\sigma)\zeta}d\sigma
$$


where $x>0$ and $\alpha$ is a real number. How does the single integral become in a double integral?



Doubt filling a step in a derivation
Working on the left-hand side, perform the change $\epsilon/x = \sigma$, in order to get


$$x^ {\alpha +1} \int \limits \_0 ^\infty \sigma^\alpha (1+\sigma)^{-1} \, \mathrm e^{-x \sigma} \, \mathrm d \sigma = x^ {\alpha +1} \mathrm e ^x \int \limits \_0 ^\infty \sigma^\alpha (1+\sigma)^{-1} \, \mathrm e^{-x (\sigma + 1)} \, \mathrm d \sigma\ .$$


Now recognize that (this is where the "extra" integral comes from)


$$(1+\sigma)^{-1} \, \mathrm e^{-x (\sigma + 1)} = x \frac {\mathrm e^{-x (\sigma + 1) \cdot \infty}} {-x(\sigma+1)} - x\frac {\mathrm e^{-x (\sigma + 1) \cdot 1}} {-x(\sigma+1)} = x \int \limits \_1 ^\infty \mathrm e^{-x (\sigma + 1) u} \ \mathrm d u \ ,$$


so the above becomes


$$x^ {\alpha +2} \mathrm e ^x \int \limits \_0 ^\infty \sigma^\alpha \left( \int \limits \_1 ^\infty \mathrm e^{-x (\sigma + 1) u} \ \mathrm d u \right) \mathrm d \sigma \ .$$


Now let $\zeta = xu$ and get


$$x^ {\alpha +1} \mathrm e ^x \int \limits \_0 ^\infty \sigma^\alpha \left( \int \limits \_x ^\infty \mathrm e^{-(\sigma + 1) \zeta} \ \mathrm d \zeta \right) \mathrm d \sigma \ .$$


Finally, use Fubini's theorem to permute the two integrals and get the right-hand side.


