I want to prove that
$$\int_{0}^{\infty} \frac{dx}{1+x^b} = \frac{\pi}{b \sin(\pi/b)} \ ,$$
where $b\in (1,\infty)$. I thought about doing it in the complex plane since the integrand is a multivalued function for general $b$, but I cannot isolate the phase shift such that this method works.
On
I want to give a second solution, which doesn't rely on special functions.
Choosing the branch cut on $(0,\infty)$ and assuming $b>1$ for the moment, we make a substitution $x^b=q, dx=\frac{1}{b}q^{\frac{1}{b}-1}dq$. The resulting integral is
$$ I(b)=\frac{1}{b}\int_0^\infty \frac{q^{1/b-1}dq}{1+q} $$
Observe that varying the argument of the integrand as $q\rightarrow q\pm i\delta$ we obtain in the limit $\delta\rightarrow 0$ $$ \underbrace{\frac{1}{b}\int_0^\infty \frac{q^{1/b-1}dq}{1+q}}_{\text{above the cut} =I(b)}\quad\text{and}\quad \underbrace{\frac{e^{2\pi i/b}}{b}\int_0^\infty \frac{q^{1/b-1}dq}{1+q}}_{\text{below the cut}= e^{2\pi i/b}I(b)} $$
Furthermore the integral has a residue at $q=-1$, which is $\text{Res}[q=-1]=\frac{-e^{i\pi/b}}{b}$.
Now choose an keyhole contour and apply residue theorem enclosing the branch cut. The contribution of the origin vanishes and we get $$ \left[1-e^{2\pi i/b}\right]I(b)=2\pi i\frac{-e^{i\pi/b}}{b}\\ \rightarrow I(b)=\frac{\pi}{b\sin(\pi/b)} $$
Let $\frac{1}{1+x^b}=u$. Then:
$$ I = \frac{1}{b}\int_{0}^{1} u^{-\frac{1}{b}}(1-u)^{\frac{1}{b}-1}\,du=\frac{\Gamma\left(1-\frac{1}{b}\right)\Gamma\left(\frac{1}{b}\right)}{b}=\frac{\pi}{b \sin\frac{\pi}{b}} $$
by Euler's Beta function and the reflection formula for the $\Gamma$ function.