The domain of convergence
We want to compute the integral \(I=\int_0^\infty dx \frac{x^\alpha}{x^2 - 2\beta x + 1 }\) for a range of real valued parameters \(\alpha\) and \(\beta\). Since the denominator is quadratic, we need to have \(\alpha<1\) so that the integral converges. Additionally, if \(\alpha\) is an integer, the integral can be evaluated by partial fractions. Therefore, we will assume that \(\alpha\) is not an integer. Furthermore, in order for the integral to converge, we also require \(-1<\alpha\). The other thing we have to check is the poles of the denominator. We first upgrade real valued parameter \(x\) to a complex number \(z\), and define \(f(z)\): \[\begin{equation} f(z)\equiv\frac{z^\alpha}{z^2 - 2\beta z +1} =\frac{z^\alpha}{(z-z_1)(z-z_2)},\tag{1} \end{equation}\] where \(z_{1,2}=\beta\pm \sqrt{\beta^2-1}\) as shown in Fig. 1.
Figure 1: The denominator has two roots: \(z_1\) and \(z_2\). The position these roots on the complex plane will depend on the value of \(\beta\). A few values of \(\beta\) are marked on the plot. Hover on the lines to see more values.
If the roots fall on the positive real x-axis, the integral will diverge. From the plot we observe that if \(\beta<1\), the roots will not be on the positive x-axis. Therefore, the integral will be well defined for \(\beta<1\) and \(-1<\alpha<1\).
The key-hole contour
Due to the \(x^\alpha\) term with non-integer \(\alpha\), the integral has a branch cut. We can take the positive x-axis as the cut.
Figure 2: Key-hole contour to evaluate the integral. The dashed lines show the possible positions of the two poles.
Using the residue theorem, we can write: \[\begin{eqnarray} \oint f(z)dz &=& 2\pi i \left( \text{Res}(f,z_1)+\text{Res}(f,z_2) \right)=2\pi i \left( \frac{z_1^\alpha}{z_1-z_2}+ \frac{z_2^\alpha}{z_2-z_1} \right)\nonumber\\ &=& \frac{\pi i}{\sqrt{\beta^2-1}} \left[ \left(\beta+\sqrt{\beta^2-1}\right)^\alpha- \left(\beta-\sqrt{\beta^2-1}\right)^\alpha \right] \tag{2}. \end{eqnarray}\] On the left hand-side, the integrals over the circles \(C_R\) and \(C_\epsilon\) vanish. We just need to figure out what happens on \(C_{1,2}\). The integral on \(C_2\) is the original integral we are looking to solve. The one on \(C_1\) is \[\begin{equation} \int_{C_1}dz f(z)=\int_{C_1}dx \frac{x^\alpha e^{i2\pi \alpha}}{x^2 - 2\beta x + 1}=-\int_\epsilon^\infty \frac{x^\alpha e^{i2\pi \alpha}}{x^2 - 2\beta x + 1}=-e^{i2\pi \alpha}I.\tag{3} \end{equation}\]
Therefore, the final result is \[\begin{equation} I=\frac{\pi i}{\sqrt{\beta^2-1}(1-e^{i2\pi \alpha})} \left[ \left(\beta+\sqrt{\beta^2-1}\right)^\alpha- \left(\beta-\sqrt{\beta^2-1}\right)^\alpha \right].\tag{4} \end{equation}\]
Various interesting cases
Let us look at a few specific cases.
\(\beta=0\) case
The roots are \(z_{1,2}=\pm i\). The corresponding integral becomes:
\[\begin{eqnarray} I&=&\int_0^\infty dx \frac{x^\alpha}{x^2 + 1 }=\frac{\pi i}{i(1-e^{i2\pi \alpha})} \left[ i^\alpha- (-i)^\alpha \right]=\frac{\pi }{1-e^{2\pi i \alpha}} \left[ e^{i\pi\alpha/2}-e^{3\pi i\alpha/2} \right]\nonumber\\ &=&\frac{\pi }{e^{-i\pi \alpha}-e^{i\pi \alpha}} \left[ e^{-i\pi\alpha/2}-e^{i\pi\alpha/2}\right]=\frac{\pi \sin(\pi\alpha/2)}{\sin(\pi\alpha)}=\frac{\pi}{2\cos(\pi\alpha/2)}\tag{5}. \end{eqnarray}\]
\(\beta=-1/\sqrt{2}\) case
The roots are \(\{z_1,z_2\}=\{e^{3\pi i/4},e^{5\pi i/4} \}\), and \(\sqrt{\beta^2-1}=1/\sqrt{2}\) The corresponding integral reads: \[\begin{eqnarray} I&=&\int_0^\infty dx \frac{x^\alpha}{x^2 + \sqrt{2} x + 1 }=\frac{i \pi \sqrt{2}}{(1-e^{2\pi i \alpha})} \left[e^{3\pi i\alpha/4} -e^{5\pi i \alpha/4} \right]=\frac{\sqrt{2}\pi \sin(\pi\alpha/4)}{\sin(\pi\alpha)}\tag{6}. \end{eqnarray}\]
\(\beta=-1\) case
This is a tricky case since the roots merge. We can either fall back onto the computation of residues with higher order poles, or we can simply approach this limit carefully by setting \(\beta=-1+\epsilon\) to get \(z_{1,2}=-1\pm \delta\) where \(\delta\equiv\sqrt{2\epsilon}\) is a small positive number. Equivalently, \(\{z_1,z_2\}=\{-e^{-i \delta},-e^{i \delta}\}\) and \(\sqrt{\beta^2-1}=\delta\). Then the integral becomes: \[\begin{eqnarray} I&=&\int_0^\infty dx \frac{x^\alpha}{x^2 + 2 x + 1 }=\frac{ \pi i}{\delta (1-e^{2\pi i\alpha})} e^{i\pi\alpha}\left[e^{ -i\delta\alpha} -e^{i \delta \alpha} \right]=\frac{\pi\alpha}{\sin(\pi\alpha)}\tag{7}. \end{eqnarray}\]
Putting it all together
Not that the complete answer is already given in Eq.(4). One could simply plug in numbers and get the answer. However, it requires surgical precision to compute the function due to the branch cut: if one is not careful enough, s/he will cross the cut, and the result will be messed up due to multi-valued nature of the functions. So let’s dive into the expression in Eq.(4) and simplify it very carefully.
\(0\leq\beta<1\) case
In this range of \(\beta\) we will have \(z_1=\beta+ i\sqrt{1-\beta^2}\equiv e^{i\theta}\) where \(\theta=\arctan\left[\frac{\sqrt{1-\beta^2}}{\beta}\right]\), and \(z_2=\beta- i\sqrt{1-\beta^2}\equiv e^{2\pi i-i\theta}\). \(z_1\) is in the first quadrant and \(z_2\) is in the fourth. Note that we defined the angle of \(z_2\) so that we don’t cross the branch cut. We can write \(I\) as \[\begin{eqnarray} I&=&\frac{\pi i}{i \sqrt{1-\beta^2}(1-e^{i2\pi \alpha})} \left[ e^{i\theta\alpha} - e^{2\pi\alpha i-i\theta\alpha} \right] =\frac{\pi }{ \sqrt{1-\beta^2}} \frac{\sin\left[\alpha(\pi-\theta) \right] }{ \sin(\pi\alpha)}\nonumber\\ &=&\frac{\pi }{ \sqrt{1-\beta^2}} \frac{\sin\left\{\alpha\left(\pi-\arctan\left[\frac{\sqrt{1-\beta^2}}{\beta}\right]\right) \right\}}{ \sin(\pi\alpha)} \tag{8}. \end{eqnarray}\]
\(-1\leq\beta<0\) case
In this range of \(\beta\) we will have \(z_1=\beta+ i\sqrt{1-\beta^2}\equiv e^{i(\pi-\theta)}\) where \(\theta=\arctan\left[\frac{\sqrt{1-\beta^2}}{|\beta|}\right]\), and \(z_2=\beta- i\sqrt{1-\beta^2}\equiv e^{i(\pi +\theta)}\). Note that we again defined the angle of \(z_2\) so that we don’t cross the branch cut. \(z_1\) is in the second quadrant and \(z_2\) is in the third. We can write \(I\) as \[\begin{eqnarray} I&=&\frac{\pi i}{i \sqrt{1-\beta^2}(1-e^{i2\pi \alpha})} e^{i\pi\alpha} \left[ e^{-i\theta\alpha} - e^{i\theta\alpha} \right] =\frac{\pi }{ \sqrt{1-\beta^2}} \frac{\sin\left[\alpha(\theta) \right] }{ \sin(\pi\alpha)}\nonumber\\ &=&\frac{\pi }{ \sqrt{1-\beta^2}} \frac{\sin\left\{\alpha\arctan\left[\frac{\sqrt{1-\beta^2}}{|\beta|}\right] \right\}}{ \sin(\pi\alpha)} \tag{9}. \end{eqnarray}\]
\(\beta<-1\) case
In this range of \(\beta\) we will have \(z_{1,2}=\beta\pm\sqrt{\beta^2-1}\), which are both negative real numbers. We can write \(I\) as \[\begin{eqnarray} I&=&\frac{\pi i}{ \sqrt{\beta^2-1}(1-e^{i2\pi \alpha})} e^{i\pi\alpha} \left[ \left(|\beta|-\sqrt{\beta^2-1}\right)^\alpha -\left(|\beta|+\sqrt{\beta^2-1}\right)^\alpha\right]\nonumber\\ &=&\frac{\pi \left[ \left(|\beta|+\sqrt{\beta^2-1}\right)^\alpha -\left(|\beta|-\sqrt{\beta^2-1}\right)^\alpha \right] }{ 2 \sqrt{\beta^2-1} \sin(\pi\alpha)} \tag{10}. \end{eqnarray}\]