Tough integral involving logarithm Let 
$$ I = \int_0^{\infty} \ln{\left(\frac{a^s + x^s}{b^s + x^s}\right)}dx$$
assuming $a>0$, $b>0$, and $s>1$. (It would be great if someone could explain why $s$ should satisfy this to guarantee convergence.)
Integration by parts yields(after an annoying limit)
$$I = s(a^s - b^s) \int_0^{\infty} \frac{x^s}{(a^s + x^s)(b^s + x^s)}dx$$
I believe this form of the integral might be ripe for contour integration, however juggling the potential branch cuts of $x^s$ along with poles at solutions of $x^s = -a^s$, and  $x^s = -b^s$ makes me think this might not be the best solution.
Perhaps there might be a better way to approach the original integral?

Let
$$ I(a,b,s)=\int_{0}^{+\infty}\log\left(\frac{a^s+x^s}{b^s+x^s}\right)\,dx.$$
In a right neighbourhood of the origin the integrand function behaves like $s\log\left(\frac{a}{b}\right)$ and such function is continuous on $\mathbb{R}^+$. Assuming $s>0$, its limit as $x\to +\infty$ is $\log(1)=0$. By setting $a^s=A, b^s=B$, we have
$$ \int_{0}^{M}\log(A+x^s)\,ds = \frac{1}{s}\int_{0}^{M^s}\log(A+t) t^{1/s}\frac{dt}{t}= \left[t^{1/s}\log(A+t)\right]_0^{M^s}-\int_{0}^{M^s}\frac{dt}{t^{1/s}(A+t)} $$
and
$$\int_{0}^{M}\log\left(\frac{A+x^s}{B+x^s}\right)\,dx=\left[t^{1/s}\log\left(\frac{A+t}{B+t}\right)\right]_0^{M^s}-\int_{0}^{M^s}\frac{(B-A)\,dt}{t^{1/s}(A+t)(B+t)}.$$
As $M\to +\infty$, the term $\left[\ldots\right]_{0}^{M^s}$ has a finite limit iff $s\geq 1$, but in order that $\frac{1}{t^{1/s}}$ is integrable in a right neighbourhood of the origin we need $s>1$. With such assumption everything is actually convergent and we may state
$$\int_{0}^{+\infty}\log\left(\frac{A+x^s}{B+x^s}\right)\,dx=(A-B)\int_{0}^{+\infty}\frac{dt}{t^{1/s}(A+t)(B+t)}.$$
Since 
$$\int_{0}^{+\infty}\frac{dt}{t^{1/s}(A+t)}=\frac{1}{A^{1/s}}\int_{0}^{+\infty}\frac{du}{u^{1/s}(1+u)}=\frac{\pi}{a\sin\frac{\pi}{s}}$$
by Euler's Beta function and the reflection formula for the $\Gamma$ function, we have:
$$\boxed{\forall s>1,\qquad \int_{0}^{+\infty}\log\left(\frac{a^s+x^s}{b^s+x^s}\right)\,dx=\frac{\pi (a-b)}{\sin\frac{\pi}{s}}.}$$
