Prove that for every real \(\alpha>1\), \[ \int_0^\infty\frac{dx}{1+x^\alpha} = \frac{\pi/\alpha}{\sin(\pi/\alpha)}. \]
Solution
Set \[ \beta\coloneqq\frac1\alpha. \] Since \(\alpha>1\), \[ 0<\beta<1. \]
1. The substitution\[ t=x^\alpha \]gives\[ \int_0^\infty\frac{dx}{1+x^\alpha} = \frac1\alpha \int_0^\infty \frac{t^{\beta-1}}{1+t}\,dt. \]
From \(t=x^\alpha\), \[ x=t^{1/\alpha}=t^\beta \] and \[ dx = \frac1\alpha t^{1/\alpha-1}\,dt = \frac1\alpha t^{\beta-1}\,dt. \] Substitute into the integral.
2. Define\[ J(\beta) \coloneqq \int_0^\infty \frac{t^{\beta-1}}{1+t}\,dt. \]Then\[ J(\beta) = \frac{\pi}{\sin(\pi\beta)}. \]
Use the branch \[ z^{\beta-1} = \exp\bigl((\beta-1)\Log z\bigr), \qquad 0<\arg z<2\pi, \] and integrate \[ \frac{z^{\beta-1}}{1+z} \] around a keyhole contour about the positive real axis.
Because \(0<\beta<1\), the integrals over the outer and inner circular arcs tend to zero as their radii tend to \(\infty\) and \(0\). The upper bank contributes \(J(\beta)\). On the lower bank the power acquires the factor \[ e^{2\pi i(\beta-1)} = e^{2\pi i\beta}, \] and the reversed orientation makes its contribution \[ -e^{2\pi i\beta}J(\beta). \]
The only enclosed pole is \(z=-1\), where the chosen argument is \(\pi\). Its residue is \[ e^{i\pi(\beta-1)}. \] Thus the residue theorem gives \[ \bigl(1-e^{2\pi i\beta}\bigr)J(\beta) = 2\pi i e^{i\pi(\beta-1)}. \] Since \[ 1-e^{2\pi i\beta} = -2ie^{i\pi\beta}\sin(\pi\beta), \] solving for \(J(\beta)\) gives \[ J(\beta) = \frac{\pi}{\sin(\pi\beta)}. \]
3. Therefore\[ \boxed{ \int_0^\infty\frac{dx}{1+x^\alpha} = \frac{\pi/\alpha}{\sin(\pi/\alpha)} }. \]
Combine steps 1 and 2 and substitute \[ \beta=\frac1\alpha. \]
4. Q.E.D.
Step 3 is the required identity.