Consider the equation \[ \int_0^\infty \frac{x^\alpha}{(1+x^2)^2}\,dx =\frac{\pi(1-\alpha)}{4\cos(\pi\alpha/2)}. \] Find the set of real \(\alpha\) for which the integral is absolutely convergent, and prove the equation for all such \(\alpha\).
Solution
The integral is absolutely convergent exactly for \[ \boxed{-1<\alpha<3}. \] For every such \(\alpha\), \[ \boxed{ \int_0^\infty \frac{x^\alpha}{(1+x^2)^2}\,dx =\frac{\pi(1-\alpha)}{4\cos(\pi\alpha/2)}.} \] At \(\alpha=1\) the right-hand side is interpreted by continuity; its value is \(1/2\).
1. The improper integral is absolutely convergent exactly when\(-1<\alpha<3\).
The integrand is nonnegative for \(x>0\), so absolute convergence is the same as convergence. Near \(0\), \[ \frac{x^\alpha}{(1+x^2)^2}\sim x^\alpha, \] and \(\int_0^1x^\alpha\,dx\) converges exactly when \(\alpha>-1\). Near infinity, \[ \frac{x^\alpha}{(1+x^2)^2}\sim x^{\alpha-4}, \] and \(\int_1^\infty x^{\alpha-4}\,dx\) converges exactly when \(\alpha-4<-1\), equivalently \(\alpha<3\). Combining the two conditions gives \(-1<\alpha<3\).
2. Suppose \(-1<\alpha<3\) and \(\alpha\notin\mathbb Z\). On\(\mathbb C\setminus[0,\infty)\) choose the branch\[ z^\alpha=\exp(\alpha(\log|z|+i\arg z)), \qquad 0<\arg z<2\pi, \]and set\[ F(z)=\frac{z^\alpha}{(1+z^2)^2}. \]For a keyhole contour around the positive real axis, the two circular arcsmake vanishing contributions as their radii tend to \(0\) and \(\infty\).
On the outer circle \(|z|=R\), \[ |F(z)|=O(R^{\alpha-4}), \] so the integral over an arc of length \(O(R)\) is \(O(R^{\alpha-3})\), which tends to \(0\) because \(\alpha<3\). On the inner circle \(|z|=\varepsilon\), \[ |F(z)|=O(\varepsilon^\alpha), \] so the integral is \(O(\varepsilon^{\alpha+1})\), which tends to \(0\) because \(\alpha>-1\).
3. If\[ I(\alpha)=\int_0^\infty\frac{x^\alpha}{(1+x^2)^2}\,dx, \]then the two straight sides of the keyhole contour contribute\[ (1-e^{2\pi i\alpha})I(\alpha). \]
On the upper side of the positive axis, \(z=x\) and \(z^\alpha=x^\alpha\), so the contribution tends to \(I(\alpha)\). On the lower side, \(\arg z\to2\pi\), hence \[ z^\alpha=e^{2\pi i\alpha}x^\alpha. \] That side is traversed from large \(x\) to small \(x\), so its contribution tends to \(-e^{2\pi i\alpha}I(\alpha)\). Their sum is the displayed expression.
4. The poles of \(F\) inside the keyhole are the double poles at \(i\) and\(-i\), and\[ \operatorname{Res}(F,i)+\operatorname{Res}(F,-i) =\frac{\alpha-1}{2}e^{\pi i\alpha}\sin\frac{\pi\alpha}{2}. \]
For \(a\in\{i,-i\}\), \[ F(z)=\frac{z^\alpha}{(z-a)^2(z+a)^2}, \] so \[ \begin{aligned} \operatorname{Res}(F,a) &=\left.\frac{d}{dz}\left(\frac{z^\alpha}{(z+a)^2}\right)\right|_{z=a}\\ &=\frac{\alpha-1}{4}a^{\alpha-3}. \end{aligned} \] For the chosen branch, \(\arg i=\pi/2\) and \(\arg(-i)=3\pi/2\). Hence \[ \begin{aligned} i^{\alpha-3}+(-i)^{\alpha-3} &=e^{\pi i(\alpha-3)/2}+e^{3\pi i(\alpha-3)/2}\\ &=2e^{\pi i\alpha}\sin\frac{\pi\alpha}{2}. \end{aligned} \] Multiplying by \((\alpha-1)/4\) gives the claimed residue sum.
5. For nonintegral \(\alpha\in(-1,3)\),\[ I(\alpha)=\frac{\pi(1-\alpha)}{4\cos(\pi\alpha/2)}. \]
By <1>2–<1>4 and the residue theorem, \[ (1-e^{2\pi i\alpha})I(\alpha) =2\pi i\cdot\frac{\alpha-1}{2}e^{\pi i\alpha} \sin\frac{\pi\alpha}{2}. \] Since \[ 1-e^{2\pi i\alpha} =-2ie^{\pi i\alpha}\sin(\pi\alpha), \] we obtain \[ \begin{aligned} I(\alpha) &=\frac{\pi(1-\alpha)\sin(\pi\alpha/2)} {2\sin(\pi\alpha)}\\ &=\frac{\pi(1-\alpha)}{4\cos(\pi\alpha/2)}, \end{aligned} \] using \(\sin(\pi\alpha)=2\sin(\pi\alpha/2)\cos(\pi\alpha/2)\).
6. The same identity holds at the integral values\(\alpha=0,1,2\) in the convergence interval.
Fix a compact interval \(K\subset(-1,3)\). There are exponents \(-1<a<b<3\) with \(K\subset[a,b]\). For \(0<x\le1\) and \(\alpha\in K\), \[ 0\le\frac{x^\alpha}{(1+x^2)^2}\le x^a, \] while for \(x\ge1\), \[ 0\le\frac{x^\alpha}{(1+x^2)^2}\le x^{b-4}. \] The two majorants are integrable. Dominated convergence therefore shows that \(I(\alpha)\) is continuous on \((-1,3)\). The right-hand side is continuous there except for an apparent singularity at \(\alpha=1\), where \[ \lim_{\alpha\to1} \frac{\pi(1-\alpha)}{4\cos(\pi\alpha/2)} =\frac12. \] Since the identity from 5 holds on the dense set of nonintegral \(\alpha\in(-1,3)\), continuity extends it to \(\alpha=0,1,2\) as well.