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?

  • $\begingroup$ The first expression should have $(-i\omega)^\alpha$ instead of $(i\omega)^\alpha$ under the outer integral. $\endgroup$
    – Mark Viola
    Commented Nov 1, 2022 at 15:58
  • $\begingroup$ Is $D^\alpha$ meant to be a fractional derivative operator? $\endgroup$
    – Mark Viola
    Commented Nov 1, 2022 at 16:12

1 Answer 1


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!


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