I am having a really difficult time trying to solve $$\int_{0}^{\infty} \frac{dx}{x^3+x+1}$$

So I know I should definitely be wanting to use residue calculus. Moreover it was asked in the context of residue thereom, so if possible that would be my only intrest

It doesn't seem similar to any of the other examples I have because I do not know a simple closed form for the poles, moreover, I dont have any sort of idea of what contour I would need to choose.

I think possibly it can be done using the complex logarithm? I know the answer is approximately $0.921763$

So any ideas/solutions on how to actually come to that final answer?


  • 1
    $\begingroup$ You may be able to force an answer out by using Partial Fractions, as the denominator does factor into a linear and a quadratic. wolframalpha.com/input/?i=factor+x%5E3%2Bx%2B1 $\endgroup$ Nov 24, 2016 at 21:40
  • $\begingroup$ You might also be able to use the trig substitution $x=\tan u$, as that can be used to get rid of some of the stuff in the denominator. $\endgroup$ Nov 24, 2016 at 21:42

2 Answers 2


$x^3+x+1$ has a single real root $\alpha$ (close to $-0.7$) and two complex conjugate roots $\beta,\overline{\beta}$.
By Vieta's formulas, $\alpha+\beta+\overline{\beta}=0$, $\alpha\beta\overline{\beta}=-1$ and $\alpha(\beta+\overline{\beta})+\beta\overline{\beta}=1$.
By partial fraction decomposition, $$ \frac{1}{x^3+x+1}=\frac{1}{(x-\alpha)(x-\beta)(x-\overline{\beta})} = \sum_{\xi\in\{\alpha,\beta,\overline{\beta}\}}\frac{\text{Res}\left(\frac{1}{x^3+x+1},x=\xi\right)}{x-\xi} \tag{1}$$ or: $$ \frac{1}{x^3+x+1} = \sum_{\xi\in\{\alpha,\beta,\overline{\beta}\}}\frac{1}{(3\xi^2+1)(x-\xi)} \tag{2} $$ so that: $$ \int_{0}^{M}\frac{dx}{x^3+x+1}=\sum_{\xi\in\{\alpha,\beta,\overline{\beta}\}}\frac{\log(M-\xi)-\log(-\xi)}{(3\xi^2+1)} \tag{3}$$ and since the sum of the residues at $\alpha,\beta,\overline{\beta}$ is zero, $$ \int_{0}^{+\infty}\frac{dx}{x^3+x+1} = \color{red}{\frac{-\log(-\alpha)}{3\alpha^2+1}-2\log\|\beta\|\cdot\text{Re}\left(\frac{1}{3\beta^2+1}\right)+2\text{Arg}(-\beta)\cdot \text{Im}\left(\frac{1}{3\beta^2+1}\right)}. \tag{4} $$ Numerically, such integral is $0.921763372185057543420173329\ldots$.
It has to be real, obviously.


Using the methods of residues, you can prove that:

$$\int_0^{\infty}\frac{x^{-p}dx}{1+x} = \frac{\pi}{\sin(\pi p)}\tag{1}$$

This then allows you to evaluate the integral of any rational function from zero to infinity, you start with writing down the partial fraction decomposition of the integrand. The idea is then that after multiplying the expression by $x^{-p}$ we obtain an analytic function of $p$, the integrals of the terms of the partial fraction expansion of the form $\frac{a}{b + x} x^{-p}$ which for $b$ not on the negative real axis can be easily evaluated using (1), possibly using an analytic continuation to allow for complex $b$. If $b$ is on the negative real axis, then this can also be handled via analytic continuation; as long as the original integral is convergent, this method will always work.

Adding up the integrals of all the terms as a function of $p$ and taking the limit of $p$ to zero will yield the desired integral.

  • $\begingroup$ Not clear why this was downvoted, clearly this is a valid method that the OP wanted to read about. $\endgroup$ Nov 25, 2016 at 18:49

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