Improper integral of piece-wise rational function I am looking to evaluate the following improper-integral
$$ \int\nolimits_0^\infty \left( \frac{1}{ 1 + x^\alpha \vert v-1 \vert^\alpha} - \frac{1}{ 1 + x^\alpha \vert v+1 \vert^\alpha} \right) \frac{ d v}{v} $$
for some positive value of $x$ and $\alpha = 3$ in view of being able to generalize this to odd integers $\alpha = 2 n-1$. 
The answer for $\alpha=3$ is known to be 
$$ \frac{2 x^3 \log(1/x)}{1-x^6} + \frac{\pi x s_3}{ (x-c_3)^2 + s_3^2} + \frac{\pi x s_3}{ (x+c_3)^2 + s_3^2} $$
where $s_3 = \sin\left(\pi/3\right)$ and $c_3 = \cos\left(\pi/3\right)$. 
I am not able to reproduce this, and would appreciate help. Thank you
 A: An integral from $0$ to $\infty$ with $\mathrm dv/v$ in it tends to be susceptible to a substitution of the form $u=\lambda v$, which leaves all of that invariant. In the present case, you can write $u=xv$ to get
$$
\int_0^\infty \left( \frac{1}{ 1 + x^\alpha \vert v-1 \vert^\alpha} - \frac{1}{ 1 + x^\alpha \vert v+1 \vert^\alpha} \right) \frac{\mathrm d v}{v}
=
\int_0^\infty \left( \frac{1}{ 1 + \vert u-x \vert^\alpha} - \frac{1}{ 1 + \vert u+x \vert^\alpha} \right) \frac{\mathrm d u}{u}\;.
$$
The integrand is even, so you can simplify things by calculating the integral from $-\infty$ to $\infty$ instead.
The denominators are of the form $z^\alpha+1$, which for odd $\alpha$ factorizes into $\prod_i(z+z_i)$, where $z_i$ are the $\alpha$-th roots of unity. So we have
$$\frac12\int_{-\infty}^\infty \left( \frac{1}{\prod_i(\vert u-x \vert+z_i)} - \frac{1}{\prod_i(\vert u+x \vert+z_i)} \right) \frac{\mathrm d u}{u}\;.$$
The rest is an exercise in partial fractions, with the two complex conjugate solutions combining into a quadratic denominator; I think you'll need to resolve the absolute values first; let me know if you want me to write it out further.
P.S.: You can further simplify things by substituting $t=u-x$ in the first term and $t=u+x$ in the second; that yields
$$\frac12\int_{-\infty}^\infty \frac{1}{\prod_i(\vert t \vert+z_i)}\left(\frac1{t+x}-\frac1{t-x} \right) \mathrm d t\;.
$$
