I am trying to show that $\int^{\infty}_1 x^s$ with $s \in \mathbb C$ converges exactly when $\mathscr Re(s)< -1$.
I have already the same result for $s \in \mathbb R$ at my disposal. I use it to show that $\mathcal Re(s) < -1 \implies \int^{\infty}_1 x^{\mathcal Re(s)} \text{ converges} \implies \int^{\infty}_1 x^{s} \text{ converges}$ because $x^{\mathcal Re(s)}=|x^s|$ and from absolute converges follows converges.
Now I need to show that $\mathcal Re(s) \ge -1 \implies \int^{\infty}_1 x^{s}$ does not converge. It is enough to show that $\mathcal Re(\int^{\infty}_1 x^{s}) = \int^{\infty}_1 x^{\mathcal Re(s)}\cos(\mathcal Im(s)\log(x))$ does not converge. I guess it follows relatively easy from the fact that $\int^{\infty}_1 x^{\mathcal Re(s)}$ does not converge and that $\cos(...)$ just adds some oscillations on top. But I am not sure how to argue this strictly.
You just follow the same procedure as for the real case: $$\int_1^N x^s\,dx=\frac{N^{s+1}-1}{s+1}$$ (for $s\ne -1$). This converges iff $N^{s+1}$ tends to a limit. But $|N^{s+1}|=N^{\Re(s)+1}$ so this can't happen when $\Re(s)>-1$ and certainly happens if $\Re(s)<-1$. This leaves $s=-1+it$ where $t\ne0$. Then $$N^{s+1}=\exp(it\log N)$$ which passes through every point on the unit circle infinitely often.