Solve an integral with a fractional power I am trying to compute the integral
$$I_{\alpha}(x):=\int_{0}^{\infty}\frac{1}{y^{1-\alpha}(x^2+y^2)(y^{\alpha}-e^{\pi \alpha \dot{\imath}})(y^{\alpha}-e^{-\pi \alpha \dot{\imath}})}dy,\quad x>0,$$
where $\alpha \in (0,1)$.
Is there a smart change of variables or an integral transformation that simplifies this ?
Mathematika cannot solve it for a general $\alpha$ with a special value of $x$, e.g. for $x=1$ (Cannot test for $x=0$ as the integral would diverge).
One could try a contour integral. But I don't know how to deal with the poles that come from the factors $y^{1-\alpha}$ or $y^{\alpha}\pm e^{\pi \alpha \dot{\imath}}$.
Of course, one could split the factors $y^{\alpha}\pm e^{\pi \alpha \dot{\imath}}$ by partial fractions and simplify the question to computing
$$J_{\alpha}(x):=\int_{0}^{\infty}\frac{1}{y^{1-\alpha}(x^2+y^2)(y^{\alpha}-e^{\pi \alpha \dot{\imath}})}dy,\quad x>0.$$
Mathematica can compute $J_{\alpha}(x)$ for any $x$ for a given $\alpha$. The answers are too complicated though to guess a formula that works for a general $\alpha$.
 A: If the complex integration is allowed, we can do the following. Let's consider the function $\displaystyle f(z)=\frac{1}{\sin\pi a}\frac{z^{a-1}}{(x^2+z^2)(z^a-e^{\pi ia})}$ and the integral along a standart keyhole contour (with the cut along the positive part of the axis X):
$$\oint f(z)dz=\frac{1}{\sin\pi a}\oint \frac{z^{a-1}}{(x^2+z^2)(z^a-e^{\pi ia})}dz$$
On the lower bank of the cut $\displaystyle f(z)=\frac{1}{\sin\pi a}\frac{z^{a-1}e^{2\pi ia}}{(x^2+z^2)(z^ae^{2\pi ia}-e^{\pi ia})}=\frac{1}{\sin\pi a}\frac{z^{a-1}}{(x^2+z^2)(z^a-e^{-\pi ia})}$, and we integrate in the negative direction. Therefore,
$$\oint f(z)dz=\frac{1}{\sin\pi a}\int_0^\infty\frac{z^{a-1}}{x^2+z^2}\Big(\frac{1}{z^a-e^{\pi ia}}-\frac{1}{z^a-e^{-\pi ia}}\Big)dz+I_r+I_R$$
where we denoted $I_r$ and $I_R$ the integrals along a big circle ($R\to\infty$) and a small circle (around $z=0; \,r\to0$) correspondingly. It is not difficult to show that $I_{r,R}\to 0$. We get
$$\oint f(z)dz=\frac{1}{\sin\pi a}\int_0^\infty\frac{z^{a-1}}{x^2+z^2}\frac{e^{\pi ia}-e^{-\pi ia}}{(z^a-e^{\pi ia})(z^a-e^{-\pi ia})}dz=2iI$$
where we denoted $I$ the desired integral. On the other hand,
$$\oint f(z)dz=2\pi i\underset{z=xe^{\pi i/2},\,xe^{3\pi i/2}, \,e^{\pi i}}{\operatorname{Res}}\frac{1}{\sin\pi a}\frac{z^{a-1}}{(x^2+z^2)(z^a-e^{\pi ia})}$$
because by means of the cut we made the function single-valued, with only three simple poles inside the contour (we turn counter-clockwise, the ange of $z$ changes from $0$ to $2\pi$).
The residues evaluation is straightforward. I got
$$\boxed{\,\,I(a,x)=\frac{\pi}{\sin\pi a}\Big(\frac{1}{a(1+x^2)}-x^{a-2}\Re\,\frac{1}{x^a-e^\frac{\pi ia}{2}}\Big)\,\,}\tag{1}$$
Further simplification is straightforward.
In the specific case of $a=1$ our initial integral is well defined (and positive):
$$I(1,x)=\int_0^\infty\frac{dy}{(x^2+y^2)(y+1)^2}$$
On the other hand, if we take in (1) $a=1-\epsilon$ and disclose the uncertainty, we will see that all divergent (at $\epsilon\to 0$) terms cancel, and we do get a finit answer.
