# $\int_0^\infty {x^a\over (x^2+1)^2} dx$ where $0<a<1$. [closed]

$\int_0^\infty {x^a\over (x^2+1)^2} dx$ 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.

## closed as off-topic by heropup, Silvia Ghinassi, Daniel Robert-Nicoud, Shailesh, user296602 Apr 22 '16 at 4:44

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Silvia Ghinassi, Daniel Robert-Nicoud, Shailesh, Community
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• Use the so-called Hankel or keyhole-contour, to exploit the fact that if we go around 0 the $a$th power of $x$ changes by $e^{2\pi i a}\not=1$, for $0<a<1$, then evaluate by residues. – paul garrett Apr 17 '16 at 20:08
• For lazy people, a method that doesn't require much computations – Count Iblis Apr 18 '16 at 0:16
• Since January 20 all but one of the questions you have posted have lacked context. Several of your posts have been closed for this reason, yet you continue to ignore all warnings that the posting of questions without any explanation of your own thoughts or effort to solve them is not allowed on this site. – heropup Apr 21 '16 at 20:21

Use the integral representation $$(x^2+1)^{-2}=\int_0^\infty d\xi\ \xi\ e^{-\xi (x^2+1)}$$ to write $$\int_0^\infty {x^a\over (x^2+1)^2} dx=\int_0^\infty d\xi\ \xi\ e^{-\xi}\int_0^\infty dx\ x^a e^{-\xi x^2}$$ $$=\int_0^\infty d\xi\ \xi\ e^{-\xi}\frac{1}{2} \xi ^{-\frac{a}{2}-\frac{1}{2}} \Gamma \left(\frac{a+1}{2}\right)=\boxed{\frac{1}{2} \Gamma \left(\frac{3-a}{2}\right) \Gamma \left(\frac{a+1}{2}\right)}$$
• Not sure if it has a 'name'. It is just the $k$-th moment of the exponential distribution with parameter $\mu$, $$\int_0^\infty d\xi\ \xi^k \mu e^{-\mu \xi}=\mu^{-k}\Gamma(1+k)\ .$$ It is extremely useful every time you have a denominator of a fraction (raised to some power), to convert it into the integral of an exponential (much more tractable). – Pierpaolo Vivo Apr 17 '16 at 20:38
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.
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}