Evaluate $\int\limits^{\infty}_{0} \frac{x^\alpha}{(1+x^2)^2}dx, \ -1 < \alpha<3.$ 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 $0<r<1.$
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 : $\begin{align} \int^{\infty}_{0} \dfrac{x^\alpha}{(1+x^2)^2}\end{align}dx=\dfrac{\dfrac{\pi(1-\alpha)i\alpha}{2}}{(1+e^{\pi i \alpha})}=\dfrac{\pi(1-\alpha)}{4\text{cos}(\pi \alpha/2)}$ 

I am providing an alternative route towards the evaluation of this integral.
Consider a change of variable $x^2=u$ then the integral can be expressed as
$$\int^{\infty}_{0}\frac{x^{\alpha}}{(1+x^2)^2}\,dx=\int^{\infty}_{0}\frac{u^{(\alpha-1)/2}}{2(1+u)^2}\,du=\frac{1}{2}\int^{\infty}_{0}\frac{u^{(\alpha-1)/2}}{(1+u)^2}\,du$$
Now using the Beta function
$$B(x,y)=\int^{\infty}_{0}\frac{u^{x-1}}{(1+u)^{x+y}}\,du$$
defined for $\Re{(x)}>0$ and $\Re{(y)}>0$ one can rewrite the integral as follows
$$\frac{1}{2}\int^{\infty}_{0}\frac{u^{(\alpha-1)/2}}{(1+u)^2}\,du=\frac{1}{2}\cdot B(\frac{\alpha+1}{2},\frac{3-\alpha}{2})$$
Now you can clearly see the restriction why $-1<\alpha<3$. Now lets evaluate the last result
$$\frac{1}{2}\cdot B(\frac{\alpha+1}{2},\frac{3-\alpha}{2})=\frac{1}{2}\cdot  B(\frac{\alpha+1}{2},1+\frac{1-\alpha}{2})=\frac{1}{2}\cdot\frac{\frac{1-\alpha}{2}}{\frac{\alpha+1}{2}+\frac{1-\alpha}{2}}\cdot B(\frac{1+\alpha}{2},\frac{1-\alpha}{2})=\frac{1-\alpha}{4}\cdot B(\frac{1+\alpha}{2},\frac{1-\alpha}{2})$$
Using the fact $$B(x,y)=\frac{\Gamma{(x)}\Gamma{(y)}}{\Gamma{(x+y)}}$$
and $$\Gamma{(x)}\Gamma{(1-x)}=\frac{\pi}{\sin{(\pi x)}}$$
then the final result reads as follows
$$\frac{1-\alpha}{4}\cdot B(\frac{1+\alpha}{2},\frac{1-\alpha}{2})=\frac{1-\alpha}{4}\cdot\frac{\Gamma{(\frac{1+\alpha}{2})}\Gamma{(\frac{1-\alpha}{2})}}{\Gamma{(1)}}=\frac{1-\alpha}{4}\cdot\frac{\pi}{\sin{(\frac{\pi(1+\alpha)}{2}})}=\frac{\pi(1-\alpha)}{4\cos{(\frac{\pi\alpha}{2}})}$$
