complex analysis - Evaluate $intlimits^{infty}_{0} frac{x^alpha}{(1+x^2)^2}dx, -1 < alpha

Evaluate $\begin{align} \int^{\infty}_{0} \dfrac{x^\alpha}{(1+x^2)^2}\end{align}dx, \ -1 < \alpha<3.$

May I verify if my solution is correct? Thank you.

Consider $\gamma_1:=\{x:-\dfrac{1}{r}\leq x \leq-r\}, \ \gamma_2:=\{re^{it}: \pi\leq t \leq 0\}, \gamma_3:\{x:r \leq x \leq\dfrac{1}{r}\}, $ $\gamma_3:= \{\frac{1}{r}e^{it}: 0\leq t\leq \pi\},$ where $0

Let $f(z)=\dfrac{z^\alpha}{(1+z^2)^2}.$

$\begin{align} \int_{\gamma_{1}}f(z)dz=\int^{-r}_{-1/r}f(z)dz= \Biggl[\begin{array}{c} z=-w,w>0 \\\ dz=-dw \end{array}\Biggr]=-\int^{r}_{1/r}\dfrac{(-w)^\alpha}{(1+(-w^2))^2}dw \end{align}$ $= \begin{align} (e^{\pi i})^{\alpha}\int^{1/r}_{r}\dfrac{w^\alpha}{(1+w^2)^2}dw\end{align}.$

$\begin{align}\int_{\gamma_{3}}f(z)dz=\int^{1/r}_{r}\dfrac{x^\alpha}{(1+x^2)^2} dx\end{align}$

$\begin{align}\left|\int_{\gamma_{4}}f(z)dz \right| \to 0, \ r \to 0\end{align},$ since $-1 < \alpha <3.$ Similarly, $\begin{align}\left|\int_{\gamma_{2}}f(z)dz \right| \to 0, \ r \to 0\end{align}$

$i$ is a double pole of $f \implies 2\pi i Res(f,i)=2\pi i\lim_{z \to i}((z-i)^2f(z))^{\prime}= \dfrac{\pi(1-\alpha)i\alpha}{2}$

By Cauchy Residue Thm, $\begin{align}\dfrac{\pi(1-\alpha)i\alpha}{2}= (1+e^{\pi i \alpha})\int^{1/r}_{r}{\dfrac{f(x)}{(1+x^2)^2}dx}+\int_{\gamma_{2} \cup \gamma_{3}}f(z)dz \end{align}.$

Letting $r \to 0,$ we have : $\begin{align} \int^{\infty}_{0} \dfrac{x^\alpha}{(1+x^2)^2}\end{align}dx=\dfrac{\dfrac{\pi(1-\alpha)i\alpha}{2}}{(1+e^{\pi i \alpha})}=\dfrac{\pi(1-\alpha)}{4\text{cos}(\pi \alpha/2)}$