TITLE: An integration question to be solved without using differentiation under the integral sign.
QUESTION [8 upvotes]: $$I(\alpha)=\int_0^1 \frac{x^\alpha-1}{\ln x}dx.$$
As the title says, if someone could solve this without using the differentiation under the integral sign technique, I would be very grateful.

REPLY [2 votes]: By substitution $u = -\ln x$, we have
$$I(\alpha)= \int_0^\infty \frac{e^{-u} - e^{-(1 + \alpha)u}}{u}\, du = \int_0^\infty \int_1^{1+\alpha} e^{-tu}\, dt\, du = \int_1^{1+\alpha} \int_0^\infty e^{-tu}\, du\, dt.$$
The interchange of double integrals is justified by the Weierstrass test, since $f(t,u) = e^{-tu}$ is continuous on $[1,1+\alpha]\times [0,\infty)$, dominated by $e^{-u}$, and $\int_0^\infty e^{-u}\, du$ is finite. Now
$$\int_1^{1+\alpha} \int_0^\infty e^{-tu}\, du\, dt = \int_1^{1 + \alpha} \frac{dt}{t} = \log(1 + \alpha),$$
so then $I(\alpha) = \log(1 + \alpha)$.