I have to compute Fourier Transform for the function $(1 - e^{-iyx})\frac{1}{x^{\alpha}}$ where $y > 0, 0.5 < \alpha < 1.5$.
My attempt: I have to compute the following integral $$I = \int_{-\infty}^{\infty} e^{itx}(1 - e^{-iyx})\frac{1}{x^{\alpha}}\, dx$$ I tried to use complex integration to deal with this problem. Firstly, let's assume t < 0, so we can consider contour $K_1$ in lower-half plane consisted of two segments $(-R, -\epsilon)$ and $(\epsilon, R)$, and two semicircles $C_{\epsilon}, C_R$ with center at $z=0$ and radius $\epsilon, R$ as well. So we have $$I_{K_1} = \int_{K_1} e^{itx}(1 - e^{-iyx})\frac{1}{x^{\alpha}} = 0,$$ as there are no singularities inside $K_1$. We can prove that
$$ \int_{C_R} e^{itx}(1 - e^{-iyx})\frac{1}{x^{\alpha}} \rightarrow 0, R \rightarrow \infty, \qquad \qquad \int_{C_\epsilon} e^{itx}(1 - e^{-iyx})\frac{1}{x^{\alpha}} \rightarrow 0, \epsilon \rightarrow 0, $$ so we have that $I = 0$ for $t < 0$. But it seems that we can use the same principe for t > 0, but in upper half plane. However it seems to be completely wrong. I guess it's because of branch cuts of $ \frac{1}{x^{\alpha}}$, but I don't know how to deal with it.
Need your help to finish this problem!