# 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

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)}

b. It does not hurt to explain the necessity that $a\in (-1,3)$.
c. Most important, in order to define $z^a$ you need to pick the branch of the logarithm which is defined in: $$\Omega=\mathbb C\smallsetminus\{it: t\le 0\}.$$ There, if $z=r\mathrm{e}^{i\vartheta}$, then $\log z=\log r+i\vartheta$, with $\vartheta\in(-\pi/2,3\pi/2)$.
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 $$\frac{1-\alpha}{4}\cdot B(\frac{1+\alpha}{2},\frac{1-\alpha}{2})=\frac{1-\alpha}{4}\cdot\frac{\Gamma{(\frac{1+\alpha}{2})}\Gamma{(\frac{1-\alpha}{2})}}{\Gamma{(1)}}=\frac{1-\alpha}{4}\cdot\frac{\pi}{\sin{(\frac{\pi(1+\alpha)}{2}})}=\frac{\pi(1-\alpha)}{4\cos{(\frac{\pi\alpha}{2}})}$$