What is the value of the definite integral of 1/(1+x^3) from 0 to infinity?

Steps to Solve

1. Factor the denominator

We start by factoring the denominator $1+x^3$. We know that $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Thus, $1 + x^3 = (1+x)(1-x+x^2)$.

2. Partial fraction decomposition

We decompose the integrand into partial fractions: $$ \frac{1}{1+x^3} = \frac{1}{(1+x)(1-x+x^2)} = \frac{A}{1+x} + \frac{Bx+C}{1-x+x^2} $$ Multiplying throughout by $1+x^3$ gives $$ 1 = A(1-x+x^2) + (Bx+C)(1+x) $$ $$ 1 = A - Ax + Ax^2 + Bx + Bx^2 + C + Cx $$ $$ 1 = (A+B)x^2 + (-A+B+C)x + (A+C) $$ Comparing the coefficients, we get: $A+B = 0$, $-A+B+C = 0$, $A+C = 1$. Thus $B=-A$ and $C=1-A$. Substituting these into the second equation: $-A - A + 1 - A = 0$, which gives $3A = 1$, so $A = \frac{1}{3}$. Then $B = -\frac{1}{3}$, and $C = 1 - \frac{1}{3} = \frac{2}{3}$.

So our partial fraction decomposition is: $$ \frac{1}{1+x^3} = \frac{1/3}{1+x} + \frac{(-1/3)x + 2/3}{1-x+x^2} = \frac{1}{3} \left( \frac{1}{1+x} + \frac{-x+2}{x^2-x+1} \right) $$

3. Integrate each term

We integrate each part separately: $$ \int_{0}^{\infty} \frac{1}{1+x} dx = \left[ \ln(1+x) \right]{0}^{\infty} = \lim{t \to \infty} \ln(1+t) - \ln(1) = \infty $$ However, we need to consider the entire expression: $$ \int_{0}^{\infty} \frac{1}{1+x^{3}} d x=\frac{1}{3} \int_{0}^{\infty} \frac{1}{1+x} d x+\frac{1}{3} \int_{0}^{\infty} \frac{-x+2}{x^{2}-x+1} d x $$ For the second integral, we have: $$ \int_{0}^{\infty} \frac{-x+2}{x^{2}-x+1} d x = -\frac{1}{2} \int_{0}^{\infty} \frac{2 x-1}{x^{2}-x+1} d x+\frac{3}{2} \int_{0}^{\infty} \frac{1}{x^{2}-x+1} d x $$ $$ =-\frac{1}{2} \left[ \ln(x^{2}-x+1) \right]{0}^{\infty}+\frac{3}{2} \int{0}^{\infty} \frac{1}{(x-1/2)^{2}+3/4} d x $$ $$ =-\frac{1}{2} \left[ \ln(x^{2}-x+1) \right]{0}^{\infty} + \frac{3}{2} \cdot \frac{2}{\sqrt{3}} \left[ \arctan\left(\frac{x-1/2}{\sqrt{3}/2}\right) \right]{0}^{\infty} $$ $$ =-\frac{1}{2} \left[ \ln(x^{2}-x+1) \right]{0}^{\infty} + \sqrt{3} \left[ \arctan\left(\frac{2x-1}{\sqrt{3}}\right) \right]{0}^{\infty} $$

4. Combine and evaluate the limits

Combining all parts, we get: $$ \frac{1}{3} \left[ \ln(1+x) \right]{0}^{\infty} - \frac{1}{6} \left[ \ln(x^2-x+1) \right]{0}^{\infty} + \frac{\sqrt{3}}{3} \left[ \arctan\left(\frac{2x-1}{\sqrt{3}}\right) \right]{0}^{\infty} $$ $$ =\frac{1}{6} \left[ 2\ln(1+x) - \ln(x^2-x+1) \right]{0}^{\infty} + \frac{\sqrt{3}}{3} \left[ \arctan\left(\frac{2x-1}{\sqrt{3}}\right) \right]{0}^{\infty} $$ $$ =\frac{1}{6} \left[ \ln\left(\frac{(1+x)^2}{x^2-x+1}\right) \right]{0}^{\infty} + \frac{\sqrt{3}}{3} \left[ \arctan\left(\frac{2x-1}{\sqrt{3}}\right) \right]_{0}^{\infty} $$ As $x \to \infty$: $\frac{(1+x)^2}{x^2-x+1} \to \frac{x^2}{x^2} = 1$, so $\ln(1) = 0$. Also $\arctan(\infty) = \frac{\pi}{2}$. At $x=0$: $\ln(\frac{(1+0)^2}{0-0+1}) = \ln(1) = 0$, and $\arctan(\frac{-1}{\sqrt{3}}) = -\frac{\pi}{6}$. Therefore: $$ \frac{1}{6} (0 - 0) + \frac{\sqrt{3}}{3} \left( \frac{\pi}{2} - \left(-\frac{\pi}{6}\right) \right) = \frac{\sqrt{3}}{3} \left( \frac{3\pi+ \pi}{6} \right) = \frac{\sqrt{3}}{3} \cdot \frac{4\pi}{6} = \frac{2\pi\sqrt{3}}{9} = \frac{2\pi}{3\sqrt{3}} $$