$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =?

Question

So guys I have this improper integral $\int_{0}^{\infty}\frac{x}{x^3+1}dx$. I checked that it converges by $ \int_{0}^{1}\frac{x}{x^3+1}dx + \int_{1}^{\infty}\frac{x}{x^3+1}dx $ and using the $\frac{c}{(x-a)^\lambda} $ and $\frac{1}{x^\lambda} $ criteria. But for finding the value I have trouble.I tried $ \int_{0}^{\infty}\frac{x}{(x+1)(x^2-x+1)}dx$ and $\int_{0}^{\frac{\pi}{2}}\frac{tg\alpha}{tg\alpha^3+1}dx$ but it gets me nowhere. Any ideas on solving it?

Answer 1

One thing that simplifies the computation is splitting the integral and substituting $y = x^{-1}$ in one part:

$$\begin{align} \int_0^\infty \frac{x}{x^3+1}\,dx &= \int_0^1\frac{x}{x^3+1}\,dx + \int_1^\infty \frac{x}{x^3+1}\,dx\\ &= \int_0^1 \frac{x}{x^3+1}\,dx + \int_1^0 \frac{y^{-1}}{y^{-3}+1}\,d(y^{-1})\\ &= \int_0^1 \frac{x}{x^3+1}\,dx + \int_0^1 \frac{1}{y^3+1}\,dy\\ &= \int_0^1 \frac{dx}{x^2 -x+1}. \end{align}$$

This can be seen to be related to an $\arctan$.

Answer 2

Continuing Daniel Fischer's answer, we have: $$\int_{0}^{1}\frac{dx}{x^2-x+1}=\int_{-1/2}^{1/2}\frac{dx}{x^2+3/4}=\frac{\sqrt{3}}{2}\cdot\frac{4}{3}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{dx}{x^2+1}=\frac{4\sqrt{3}}{3}\arctan\frac{1}{\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}.$$

Answer 3

If $I=\int_{0}^{\infty}\frac{x}{x^3+1}dx$ then $2I=\int_{0}^{\infty}\frac{2x-x^2+x^2}{x^3+1}dx$=$\int_{0}^{\infty}\frac{2x-x^2}{x^3+1}dx$+$\int_{0}^{\infty}\frac{x^2}{x^3+1}dx.$ The secod itegral is easy and can be solved by substituting $u=x^3+1$. For the first write $$\int_{0}^{\infty}\frac{2x-x^2}{x^3+1}dx=\int_{0}^{\infty}\frac{(x+1)-(x^2-x+1)}{(x+1)(x^2-x+1)}dx=\int_{0}^{\infty}\frac{1}{x^2-x+1}dx-\int_{0}^{\infty}\frac{1}{x+1}dx$$ but for the $\int_{0}^{\infty}\frac{1}{x^2-x+1}dx$=$\int_{0}^{\infty}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}dx$ which is of the form $\int \frac{1}{u^2+a^2}du=\frac{1}{a}\arctan(\frac{u}{a})+c.$

Answer 4

The other answers have already shown elementary methods, so I will just add that we actually have $$ \int_0^{\infty} \frac{x^a \, dx}{1+x^b} = \frac{\pi}{b\sin\left(\frac{\pi(a+1)}{b}\right)}$$ for all $a, b$ such that the integral converges. Unfortunately this is not easy to show without complex analysis. It requires knowledge of the Beta function.