$\int_0^\infty {x^a\over (x^2+1)^2} dx$ where $0$\int_0^\infty {x^a\over (x^2+1)^2} dx$ where $0<a<1$.
I know I can use partial fraction decomposition to obtain two different integrals, but I'm not sure how to integrate them. Any solutions or hints are greatly appreciated.
 A: Use the integral representation
$$
(x^2+1)^{-2}=\int_0^\infty d\xi\ \xi\ e^{-\xi (x^2+1)}
$$
to write
$$
\int_0^\infty {x^a\over (x^2+1)^2} dx=\int_0^\infty d\xi\ \xi\ e^{-\xi}\int_0^\infty dx\ x^a e^{-\xi x^2}
$$
$$
=\int_0^\infty d\xi\ \xi\ e^{-\xi}\frac{1}{2} \xi ^{-\frac{a}{2}-\frac{1}{2}} \Gamma \left(\frac{a+1}{2}\right)=\boxed{\frac{1}{2} \Gamma \left(\frac{3-a}{2}\right) \Gamma \left(\frac{a+1}{2}\right)}
$$
A: The function $z^{a}$ is holomorphic in the slitted plane $\mathbb{C}\setminus[0,\infty)$, with a jump discontinuity on the slit that is equal to
$$
       \lim_{\epsilon\downarrow 0} \{(x+i\epsilon)^{a}-(x-i\epsilon)^{a}\}
      = x^{a}(1-e^{2\pi i a})
$$
Therefore,
$$
    \int_{0}^{\infty}\frac{x^a}{(1+x^2)^2}dx=\lim_{R\rightarrow\infty}\frac{1}{1-e^{2\pi ia}}\int_{C_{R}}\frac{z^{\alpha}}{(1+z^2)^2}dz,
$$
where $C_{R}$ is the positively-oriented contour starting at $R+i0$ on the real axis, circling counter-clockwise on the circle of radius $R$ centered at the origin until you reach $R-i0$, and then following the real axis from below until you reach the origin, and finally following the real axis from above back to $R+i0$. For large $R$, the contour encloses singularities at $z = \pm i$, and, by Cauchy's representation,
\begin{align}
    \frac{1}{2\pi i}\int_{C_{R}}\frac{z^{\alpha}}{(z-i)^2(z+i)^2}dz
     & = \left.\frac{d}{dz}\frac{z^a}{(z+i)^2}\right|_{z=i}+\left.\frac{d}{dz}\frac{z^{a}}{(z-i)^2}\right|_{z=-i}.
\end{align}
Just number crunching and simplification after that.
A: If you know about the $\text{B}$ and $\Gamma$ functions, you can first let $x^2=u$ and then $t=\frac1{1+u}$ to get
$$\begin{align}\int_9^{\infty}\frac{x^a}{(x^2+1)^2}dx&=
\frac12\int_0^{\infty}\frac{u^{\frac{a-1}2}}{(1+u)^2}du
=\frac12\int_0^1(1-t)^{\frac{a-1}2}t^{\frac{1-a}2}dt\\
&=\frac12\text{B}\left(\frac{a+1}2,\frac{3-a}2\right)
=\frac12\frac{\Gamma\left(\frac{a+1}2\right)\Gamma\left(\frac{3-a}2\right)}{\Gamma\left(2\right)}\\
&=\frac12\Gamma\left(\frac{a+1}2\right)\left(\frac{1-a}2\right)\Gamma\left(\frac{1-a}2\right)\\
&=\frac{1-a}4\cdot\frac{\pi}{\sin\left(\frac{\pi(1-a)}2\right)}
=\frac{\pi(1-a)}{4\cos\left(\frac{\pi a}2\right)}\end{align}$$
