An integration question to be solved without using differentiation under the integral sign. $$I(\alpha)=\int_0^1 \frac{x^\alpha-1}{\ln x}dx.$$
As the title says, if someone could solve this without using the differentiation under the integral sign technique, I would be very grateful.
 A: We have
$$x^a = e^{a\log x} = \sum_{k=0}^{\infty} \dfrac{(a \log x)^k}{k!}$$
Hence,
$$\dfrac{x^a-1}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k \log^{k-1}(x)}{k!}$$
Make use of the  fact that
$$\int_0^1 \log^m(x) dx = (-1)^m m!$$
Hence, we obtain
$$\dfrac{x^a-1}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} (-1)^{k-1} (k-1)! = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{a^k}k =  \log(1+a)$$
Note that the above derivation is only valid when $a \in (-1,1]$. For $a>1$, it is easy to show by similar method that
$$I(a) - I(a-1) = \log(1+1/a)$$
Since $I(a) = \log(1+a)$ on the interval $[0,1]$, we can now conclude that
$$I(a) = \log(1+a)$$
for all $a > -1$.
A: 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 $I(\alpha) = \log(1 + \alpha)$.
