Let $L^2((0,1),dx)$ be the set of square integrable functions with domain $(0,1)$. Then the question is
$f_\alpha \in L^2((0,1),dx)$ where $f_\alpha:(0,1) \subset \mathbb{R} \to \mathbb{C}$ is defined by $f_\alpha(x)=x^\alpha$ for what values of $\alpha \in \mathbb{C}$?
My goal: First, the set $ L^2((0,1),dx)$ is the set of all integrable functions $f$ such that
$$\int_0^1\vert f(x)\vert^2 dx < \infty$$
Is this definition that I'm using correct?
Then, I show that we have (define $\alpha:= Re(\alpha)+iIm(\alpha) \in \mathbb{C}$) $$x^\alpha = x^{Re(\alpha)}x^{iIm(\alpha)} = x^{Re(\alpha)}e^{iIm(\alpha)\ln(x)}$$
And $\ln(x) \in \mathbb{R}$ for all $x \in (0,1)$. So we conclude that it is for all $\alpha \in \mathbb{C}$ because
$$\tag{$Re(\alpha) \neq 0$}\int_0^1\vert x^\alpha\vert^2 dx = \int_0^1\vert x^{Re(\alpha)}\vert^2 dx = \frac{1}{2Re(\alpha)} < \infty$$
And for $\Re(\alpha)=0$ we'll have $1$. Is this correct? It has an method that is general for this type of question? Can be made more rigorous (supposing that it is a little bit at least)?
We have ${ \left|x^\alpha\right|}^2 ={|x|}^{2\,\Re(\alpha)}$. Since $x\in[0,1]$, we can do away with the absolute value.
Now, for $2\,\Re(\alpha)\neq -1$ we have an antiderivative in
$$\frac{x^{2\,\Re(\alpha)+1}}{2\,\Re(\alpha)+1}$$
At $x=1$, the numerator is always well defined, but as $x\to0$, if $2\,\Re(\alpha)+1<0$, that is, if
$$\Re(\alpha)<-\frac12,$$
then the numerator diverges to infinity.
Finally, when $2\,\Re(\alpha)=-1 \iff \Re(\alpha)=-\frac12$, the antiderivative is $\ln(x)$. Here, again, as $x\to0$ the result diverges to infinity.
It follows that the condition is $\Re(\alpha)>-\frac12$.