- Use contour integration to evaluate
\[ \int _ { 0 } ^ { + \infty } { \frac { d x } { x ^ { 1 / 3 } ( 1 + x ) } } . \]
Hint
Use a keyhole contour around the positive real axis, with inner radius \(\varepsilon\) and outer radius \(R\).
Solution
The value is \(\boxed{2\pi/\sqrt3}\).
1. The integral converges absolutely at both endpoints.
For \(0<x\leq1\), the positive integrand is at most \(x^{-1/3}\), whose integral at zero is finite. For \(x\geq1\), it is at most \(x^{-4/3}\), integrable at infinity. Write \(I=\int_0^\infty x^{-1/3}/(1+x)\,dx\).
2. The logarithm branch gives a keyhole residue identity.
Use \(0<\arg z<2\pi\) and set \[ F(z)=\frac{\exp(-\operatorname{Log}z/3)}{1+z}. \] For fixed \(0<\varepsilon<1<R\), first integrate over the positive boundary of \(\{\varepsilon<|z|<R,\ \delta<\arg z<2\pi-\delta\}\), where \(0<\delta<\pi/2\). The branch is holomorphic near this contour. Its only enclosed pole is \(-1\), simple, with residue \(e^{-i\pi/3}\). The residue theorem gives the contour integral \(2\pi i e^{-i\pi/3}\) [@SS03].
Let \(\delta\downarrow0\) with the radii fixed. On each radial segment the integrand converges uniformly, since \(x\) stays in the compact interval \([\varepsilon,R]\) and the limiting denominator is nonzero. The upper bank, traversed outward, contributes \(I_{\varepsilon,R}\). The lower bank has fractional-power value \(e^{-2\pi i/3}x^{-1/3}\) and is traversed inward, so contributes \(-e^{-2\pi i/3}I_{\varepsilon,R}\). The arc integrals also have limits by boundedness on their fixed circles. Hence \[ (1-e^{-2\pi i/3})I_{\varepsilon,R} +\text{outer arc integral}+\text{inner arc integral} =2\pi i e^{-i\pi/3}. \]
3. The arcs disappear and the phase factors determine \(I\).
The respective arc integrals have moduli bounded by \[ \frac{2\pi R^{2/3}}{R-1}\longrightarrow0, \qquad \frac{2\pi\varepsilon^{2/3}}{1-\varepsilon}\longrightarrow0. \] These follow from \(|z^{-1/3}|=|z|^{-1/3}\), the arc lengths, and \(|1+z|\geq\bigl||z|-1\bigr|\). Letting \(R\to\infty\) and \(\varepsilon\downarrow0\) in step 2 therefore yields \[ (1-e^{-2\pi i/3})I=2\pi i e^{-i\pi/3}. \] Since \(1-e^{-2\pi i/3}=2i e^{-i\pi/3}\sin(\pi/3)\), division gives \(I=\pi/\sin(\pi/3)=2\pi/\sqrt3\).