Let us find all values of $\alpha$ such that the following improper integral converges:
$$I(\alpha):=\int_{1}^{\infty} \frac{ e^{\dot{\imath}(x^2-2x)}}{x^{\alpha}}\,dx$$.
When $\alpha>1$, $I(\alpha)$ is absolutely convergent.
When $-1<\alpha\leq 1$, the integral converges by the Dirichlet test, since in that case:
$$I(\alpha):=c+e^{-\dot{\imath}}\int_{2}^{\infty} \frac{ e^{\dot{\imath}(x-1)^{2}}}{x^{\alpha}}\,dx$$ where $$c=\int_{1}^{2} \frac{ e^{\dot{\imath}(x-1)^{2}}}{x^{\alpha}}\,dx,$$ and changing variables implies
$$\int_{2}^{\infty} \frac{ e^{\dot{\imath}(x-1)^{2}}}{x^{\alpha}}\,dx= \frac{1}{2}\int_{2}^{\infty} \frac{ e^{\dot{\imath}(x-1)^{2}}2(x-1)}{x^{\alpha}(x-1)}\,dx\\ =\frac{1}{2}\int_{1}^{\infty} \frac{ e^{\dot{\imath}y}}{(\sqrt{y}+1)^{\alpha}\sqrt{y}}\,dy$$ which converges by the Dirichlet test. Notice that $\frac{1}{(\sqrt{y}+1)^{\alpha}\sqrt{y}}\rightarrow 0$ when $y\rightarrow \infty$ for all $\alpha>-1$.
When $\alpha=-1$, $$I(-1)=\frac{e^{-\dot{\imath}}}{2}\left(\int_{1}^{\infty}2(x-1) e^{\dot{\imath}(x-1)^{2}}dx+2\int_{1}^{\infty} e^{\dot{\imath}(x-1)^{2}}dx \right)$$
The second integral converges by the Dirichlet test as before ($\alpha=0$ here). But, the first integral $\int_{1}^{\infty}2(x-1) e^{\dot{\imath}(x-1)^{2}}dx= \int_{0}^{\infty} e^{\dot{\imath}x}dx$ which diverges.
If $\alpha<-1$ we can show that $I(\alpha)$ diverges in a fashion similar to the case $\alpha=-1$; we integrate by parts as many times as necessary we get the divergent integral $\int_{0}^{\infty} e^{\dot{\imath}x}dx$ plus other integrals that converge by the Dirichlet test.
I would appreciate it if any expert reviewed my solution.
Thanks a lot.