# 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?

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

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!