What is the complex conjugate of this Fourier based function with complex power function? 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?
 A: To answer the first question, we begin with the $\alpha$'th fractional derivative of $f$, where $f\in \mathbb{R}$ and $\alpha\in \mathbb{R}$.  This can be expressed as
$$D^\alpha f(t)=\frac1{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty (-i\omega)f(\tau) e^{-i\omega(t-\tau)}\,d\tau\,d\omega\tag1$$
Now, we use the definition $z^w=e^{w\log(z)}$ with $-\pi<\arg(z)\le \pi$.  Then certainly we can assert that
$$(-i\omega)^\alpha =|\omega|^\alpha e^{-i\alpha\text{sgn}(\omega)\pi/2}\tag2$$
Using $(2)$ in $(1)$ reveals
$$\begin{align}
\require{cancel}
D^\alpha f(t)&=\frac1{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty |\omega|^\alpha e^{-i\alpha\text{sgn}(\omega)\pi/2}f(\tau) e^{-i\omega(t-\tau)}\,d\tau\,d\omega\\\\
&=\frac1{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty |\omega|^\alpha e^{-i[\alpha\text{sgn}(\omega)\pi/2+\omega(t-\tau)]}f(\tau) \,d\tau\,d\omega\\\\
&=\frac1{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \underbrace{|\omega|^\alpha \cos\left([\alpha\text{sgn}(\omega)\pi/2+\omega(t-\tau)]\right)}_{\text{an even function of}\,\,\omega}f(\tau) \,d\tau\,d\omega\\\\
&-i\cancelto{0}{\frac1{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \underbrace{|\omega|^\alpha \sin\left([\alpha\text{sgn}(\omega)\pi/2+\omega(t-\tau)]\right)}_{\text{an odd function of}\,\,\omega}f(\tau) \,d\tau\,d\omega}\\\\
&=\underbrace{\frac1{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty |\omega|^\alpha \cos\left([\alpha\text{sgn}(\omega)\pi/2+\omega(t-\tau)]\right)f(\tau) \,d\tau\,d\omega}_{\text{a purely real quantity}}\\\\
\end{align}$$
as was to be shown!  That is the representation in $(1)$ is purely real.
To answer second question, we appeal to $(2)$.  It is easy to see that
$$\begin{align}
\left((-i\omega)^\alpha\right)^*&=\left(|\omega|^\alpha e^{-i\alpha\text{sgn}(\omega)\pi/2}\right)^*\\\\
&=|\omega|^\alpha e^{+i\alpha\text{sgn}(\omega)\pi/2}\\\\
&=\left(|-\omega|^\alpha e^{-i\alpha\text{sgn}(-\omega)\pi/2}\right)\\\\
&=(+i\omega)^\alpha)
\end{align}$$
We conclude that $\left((-i\omega)^\alpha\right)^*=(+i\omega)^\alpha)$.
And we are done!
