I would like to compute the following integral:
$$I_n = \lim_{\epsilon\rightarrow 0^+}\int_{-\infty-i\epsilon}^{\infty-i\epsilon} \frac{d\omega}{\omega^n} e^{i\omega}$$
where $n$ is a positive half-odd integer. Observe that the integration is just below the real $\omega$-axis.
In the case that $n$ is a positive integer, I can easily do the integral by closing the contour in the upper half complex $\omega$-plane (which vanishes thanks to the exponential). Then Cauchy-residue theorem implies
$$I_n = \frac{2\pi i^n}{\Gamma(n)}\,, \qquad n\in \text{positive integers.}$$
However, if $n$ is a half-odd integer, I can do a change of variables $\omega = x^2$ to obtain $I_n = \int_C\frac{dx}{x^{2n-1}}e^{i x^2}$. But now, $2n-1$ is even, and there's no chance of using Cauchy's theorem! What should I do?
I believe your formula holds for all real numbers larger than 1. To prove it, we can use the Fourier transform (yes please!), and a good deal of computations.
Let $\alpha>1$ be a real number. Write $\omega=\xi-i\varepsilon$ with $\xi\in\mathbb R$ and $\varepsilon>0$. $$\int_{-\infty-i\epsilon}^{\infty-i\epsilon} \frac{e^{i\omega}}{\omega^\alpha}d\omega =e^{\varepsilon}\int_{\mathbb R} \frac{e^{i\xi}}{(\xi-i\varepsilon)^\alpha} d\xi=e^\varepsilon i^\alpha\int_{\mathbb R}\frac{e^{i\xi}}{(\varepsilon+i\xi)^\alpha}d\xi\tag{1}$$ Now, $$\begin{split}\int_0^{+\infty}e^{-2i\pi\xi t}e^{-2\pi \varepsilon t}t^{\alpha-1} dt &=\sum_{k\geq 0}\int_0^{+\infty}\frac{(-2i\pi\xi t)^k}{k!}e^{-2\pi\varepsilon t}t^{\alpha -1}dt\\ &=\sum_{k\geq 0}\frac{(-2i\pi\xi)^k}{k!}\int_0^{+\infty}e^{-2\pi\varepsilon t}t^{k+\alpha -1}dt \\ &= \sum_{k\geq 0}\frac{(-2i\pi\xi)^k}{k!}\frac{\Gamma(k+\alpha)}{(2\pi\varepsilon)^{k+\alpha}}\\ &= \frac 1 {(2\pi\varepsilon)^\alpha}\sum_{k\geq 0}\frac{(-i\xi)^k}{k!}\cdot\frac{\alpha(\alpha+1)(\alpha+2)...(\alpha+k-1)\Gamma(\alpha)}{\varepsilon^{k}}\\ &= \frac {\Gamma(\alpha)} {(2\pi\varepsilon)^\alpha}\sum_{k\geq 0}\left(\frac{i\xi}{\varepsilon}\right)^k\cdot\frac{(-\alpha)(-\alpha-1)(-\alpha-2)...(-\alpha-k+1)}{k!}\\ &= \frac {\Gamma(\alpha)} {(2\pi\varepsilon)^\alpha}\left(1+\frac{i\xi}\varepsilon\right)^{-\alpha}\\ &= \frac{\Gamma(\alpha)}{(2\pi)^\alpha}\cdot\frac 1{(\varepsilon+i \xi)^{\alpha}} \end{split}$$ Therefore the Fourier transform of $t\mapsto \frac{(2\pi)^\alpha}{\Gamma(\alpha)}e^{-2\pi \varepsilon t}t^{\alpha-1}H(t)$, where $H$ is the Heaviside step function, is $\xi\mapsto\frac 1 {(\varepsilon+i\xi)^\alpha}$ is . By the Fourier inversion formula, $$\int_{\mathbb R}\frac{e^{2i\pi\xi t}}{(\varepsilon+i\xi)^\alpha}d\xi=\frac{(2\pi)^\alpha}{\Gamma(\alpha)}e^{-2\pi\varepsilon t}t^{\alpha-1}H(t)$$ Evaluate at $t=\frac 1 {2\pi}$: $$\int_{\mathbb R}\frac{e^{i\xi}}{(\varepsilon+i\xi)^\alpha}d\xi=\frac{2\pi}{\Gamma(\alpha)}e^{-\varepsilon}$$ Matching with $(1)$, this yields $$\boxed{\int_{-\infty-i\epsilon}^{\infty-i\epsilon} \frac{e^{i\omega}}{\omega^\alpha}d\omega =\frac{2\pi}{\Gamma(\alpha)} i^\alpha=I_\alpha}$$