Question 2.3. Evaluate \(\textstyle \int _ { 0 } ^ { \infty } { \frac { x ^ { - { \frac { 1 } { 3 } } } } { 1 + x } } d x .\)
Solution
The value is \(\boxed{2\pi/\sqrt3}\).
1. The integral converges, and a cubic substitution makes it rational.
For \(0<x\leq1\), the integrand is at most \(x^{-1/3}\), whose integral is \(3/2\) on \((0,1]\). For \(x\geq1\), it is at most \(x^{-4/3}\), whose integral is \(3\) on \([1,\infty)\). Hence the positive improper integral converges. The substitution \(x=t^3\), first on compact intervals away from zero and then passing to the convergent endpoint limits, gives \[ I=\int_0^\infty\frac{3t}{1+t^3}\,dt. \] Factoring \(1+t^3=(t+1)(t^2-t+1)\) yields the identity \[ \frac{3t}{1+t^3}=-\frac1{t+1} +\frac{t+1}{t^2-t+1}. \] Indeed, multiplication by the denominator gives \(-(t^2-t+1)+(t+1)^2=3t\).
2. A primitive and its endpoint limits give the asserted value.
Since \(t^2-t+1=(t-1/2)^2+3/4>0\), differentiation verifies that a primitive for \(t\geq0\) is \[ H(t)=-\log(t+1)+\frac12\log(t^2-t+1) +\sqrt3\arctan\frac{2t-1}{\sqrt3}. \] The two logarithmic derivatives combine with the arctangent derivative to give \(-1/(t+1)+(t+1)/(t^2-t+1)\). At zero, \(H(0)=-\sqrt3\pi/6\). At infinity, the logarithmic contribution is \[ \frac12\log\frac{t^2-t+1}{(t+1)^2}\longrightarrow0, \] and the arctangent tends to \(\pi/2\). Thus \[ I=\lim_{t\to\infty}H(t)-H(0) =\frac{\sqrt3\pi}{2}+\frac{\sqrt3\pi}{6} =\frac{2\pi}{\sqrt3}. \]