For $-1 < \alpha < 1$ real, the below integral is well-known and an elementary exercise in complex analysis: \begin{align*} f(\alpha) \overset{\text{def}}{=} \displaystyle \int_0^\infty \frac{x^\alpha}{1 + x^2} \ \mathrm{d}x = \frac{\pi}{2} \sec \left(\frac{\pi}{2} \alpha\right ) \end{align*} However, $f$ can be extended as a complex function. In particular, over the region $D = \{\alpha \in \mathbb{C} : -1 < \operatorname{Re}(\alpha) < 1\}$ the above still holds, and I am reasonably sure $f$ blows up outside $\overline{D}$. I was interested in the behaviour of $f$ over $\overline{D} \setminus D$ (i.e. when $\operatorname{Re}(\alpha) = \pm 1$).
It is easily seen that if $\alpha = \pm 1$ then $f$ blows up, but e.g. Mathematica claims the above formula holds if $\alpha = 1 + i$. However, Mathematica also won't give an output if I substitute $x = \tan\theta$ into the integral for $\alpha = 1 + i$, and claims it is not convergent in a slightly older version, so I am hesitant in accepting its answer.
I would like a solution that carefully analyses and determines the convergence of the integral over $\overline{D}\setminus D$.
First of all, your integral is never absolutely convergent because for all $x > 0$ and all $\alpha$ of real part $\pm 1$, $|x^\alpha| = x^{\Re(\alpha)} = x^{\pm 1}$. Therefore, we have to give up Lebesgue convergence and look for Riemann convergence in the sens, $$ \int_0^\infty \frac{x^\alpha}{1 + x^2}dx = \lim_{a \rightarrow 0, b \rightarrow +\infty} \int_a^b \frac{x^\alpha}{1 + x^2}dx, $$ whenever it exists. Let $\alpha$ be such that $\Re(\alpha) = \pm 1$. First of all, we have for all $0 < a < b$, \begin{align*} \int_a^b \frac{x^\alpha}{1 + x^2}dx & = \int_{1/a}^{1/b} \frac{y^{-\alpha}}{1 + y^{-2}}\left(-\frac{dy}{y^2}\right) \textrm{ with } y = \frac{1}{x},\\ & = \int_{1/b}^{1/a} \frac{y^{-\alpha}}{1 + y^2}dy.\\ \end{align*} We deduce from this that the set $S \subset \partial D$ of complex numbers of real part $\pm 1$ such that the integral converges is stable by $\alpha \rightarrow -\alpha$ with $f(-\alpha) = f(\alpha)$. The fact that for all $x > 0$, $\overline{x^\alpha} = x^{\overline{\alpha}}$ means that it is also stable by $\alpha \rightarrow \overline{\alpha}$ with $\overline{f(\alpha)} = f(\alpha)$ (those equality are also true in general for $\frac{\pi}{2}\sec\left(\frac{\pi}{2}\alpha\right)$).
Therefore, it suffices to study the case $\alpha = 1 + it$ with $t > 0$. The integrand can we rewritten as, $$ \frac{x^\alpha}{1 + x^2} = \frac{x}{1 + x^2}e^{it\ln(x)}. $$ Clearly, $0$ is no more a problem for the convergence thus the convergence of the integral on $]0,+\infty[$ is equivalent to the convergence of $\displaystyle \int_1^b \frac{x}{1 + x^2}e^{it\ln(x)}dx$ when $b \rightarrow +\infty$ and we have for all $b > 1$, \begin{align*} \int_1^b \frac{x}{1 + x^2}e^{it\ln(x)}dx & = \int_0^{\ln(b)} \frac{e^y}{1 + e^{2y}}e^{ity}(e^ydy) \textrm{ with } x = e^y,\\ & = \int_0^{\ln(b)} \frac{e^{2y}}{1 + e^{2y}}e^{ity}dy\\ & = \int_0^{\ln(b)} e^{ity}dy - \int_0^{\ln(b)} \frac{1}{1 + e^{2y}}e^{ity}dy\\ & = \frac{e^{it\ln(b)} - 1}{it} - \int_0^{\ln(b)} \frac{1}{1 + e^{2y}}e^{ity}dy. \end{align*} The second integral converges (it absolutely converges actually) but $\frac{e^{it\ln(b)} - 1}{it}$ diverges (while being bounded) when $b \rightarrow +\infty$.
It means that in general, the quantity $$ \int_a^b \frac{x^{\pm 1 + it}}{1 + x^2}dx, $$ always diverges ($S = \emptyset$) when $a \rightarrow 0$ and $b \rightarrow +\infty$ but it remains bounded when $t \neq 0$ (and only when $t \neq 0$).