Let $\alpha \in \mathbb{R}$, $f \in L^2(\mathbb{R})$ and $z^\alpha := e^{\alpha \text{Ln}(z)}$ (the principle value of the complex power function), where $\text{Ln}(z) := \ln|z| +i\text{arg}(z)$, with arg$(z) \in (-\pi, \pi]$. Notice that this power function has a branch cut on the negative real axis. Now consider
\begin{align} g(t) = D^\alpha f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega e^{-i\omega t}(i\omega)^\alpha\int_{-\infty}^{\infty} d\tau e^{i\omega \tau}f(\tau) \end{align} My first question: is this function real or not? I thought I proved this by taking the complex conjugate and then making the substitution $\omega \to -\omega$. Is this a valied proof?
My second question $((i\omega)^\alpha)^* \stackrel{?}{=} (-i\omega)^\alpha$ (where the $*$ denotes the complex conjugate), since I know that $(-i\omega)^\alpha = (-1)^\alpha (i\omega)^\alpha e^{2\pi i\alpha N_+}$ where $N_+ = \begin{cases} &-1, \omega > 0\\ &0, \omega \leq 0 \end{cases}$. But this implies that for $k \in L^2(\mathbb{R})$
\begin{align} \int_{-\infty}^\infty dt D^\alpha( f(t)) k(t) \neq (-1)^\alpha e^{2\pi i\alpha N_+}\int_{-\infty}^\infty dt f(t) D^\alpha k(t) \end{align} where the RHS was obtained by writing out $D^\alpha$ and then substituting $\omega \to -\omega$ and swapping order of integration. But if g(t) is real then the left side also has to be real but it is clearly not. So what am I doing wrong?