Consider 2 real functions $f, g: \mathbb{R}\to\mathbb{R}$ and $\alpha \in \mathbb{R}$. First I thought that the following equality was true \begin{align} &\int_{-\infty}^{\infty}dtg(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) \\ = &(-1)^\alpha \int_{-\infty}^{\infty} d\tau f(\tau)\frac{1}{2\pi}\int_{-\infty}^{\infty} d\omega e^{-i\omega \tau} (i\omega)^\alpha\int_{-\infty}^{\infty} dt e^{i\omega t} g(t) \\ \end{align} where we made the substitution $\omega \to -\omega$.
However on further inspection I noticed that the left side is real (can be checked by taking the complex conjugate) but the right hand side is not (since $(-1)^\alpha = e^{i\pi\alpha}$). So my question is what am I missing?
Edit:
I think the fault is when I do the substitution $(-1)^\alpha (i\omega)^\alpha \stackrel{?}{=} (-i\omega)^\alpha$. But then how do I isolate the -?