How to integrate: $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}dx$

Question

I want to evaluate $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}$. The lecture only provided me with a formula for $\int_{-\infty}^{\infty}dx \frac{A}{B}$ where $A,B$ are polynomials and $B$ does not have real zeros. Unfortunately, in the given case $B$ has a zero at $z=-1$ and is not even. Is there a straight forward way to solve this in terms of complex analysis?

Answer 1

Yes there is , integrate the following function

$$f(z) = \frac{\log(z)}{z^3+z^2+z+1}$$

Around a key-hole contour. Due to the evaluation of the lines on the positive x-axis the $\log$ will cancel. What is remaining is the evaluation of the residues of the three poles.

More explanation

we have three poles $\pm i , -1$. Integrate around the following contour

$$\int_{r}^{\varepsilon}\frac{\log|x|}{x^3+x^2+x+1}dx +\int^{r}_{\varepsilon}\frac{\log|x|+2\pi i }{x^3+x^2+x+1}dx+ \int_{\gamma_r}f(z) dz +\int_{\gamma_{\varepsilon}}f(z) dz = 2\pi i \left(\mathrm{Res}(f,\pm i)+\mathrm{Res}(f,-1)\right)$$ The evaluation of the residues

$$\mathrm{Res}(f,i) = \frac{\log(i)}{3\cdot i^2+2\cdot i +1} = \frac{i\pi/2}{2i-2} $$ $$\mathrm{Res}(f,-i) = \frac{\log(-i)}{3\cdot i^2-2\cdot i +1} = \frac{3i\pi/2}{-2i-2}$$ $$\mathrm{Res}(f,-1) = \frac{\log(-1)}{3-2 +1} = \frac{\pi i}{2}$$

By summing the residues and taking $r\to \infty , \varepsilon \to 0$. Note that the complex logarithm is easily bounded hence the integrals on the smaller and bigger circles vanish and we get

$$\int^\infty_0 \frac{\log(x)}{x^3+x^2+x+1}dx -\int^ \infty_0 \frac{\log(x)+2\pi i }{x^3+x^2+x+1}dx = -\frac{\pi}{2}i$$

Hence we get

$$\int^\infty_0 \frac{\log(x)}{x^3+x^2+x+1}dx = \frac{\pi}{4}$$

Answer 2

Hint:

Factor $$x^3+x^2+x+1=(x+1)(x^2+1)$$ and use partial fractions decomposition.

Answer 3

$$\frac{1}{x^3+x^2+x+1}=\frac{1}{(x+1)(x^2+1)}=\frac{1}{2}\left(\frac{1}{1+x}-\frac{x}{x^2+1}+\frac{1}{x^2+1}\right)$$

Answer 4

$$\int_{0}^{+\infty}\frac{x-1}{x^4-1}\,dx = \int_{0}^{1}\frac{1-x}{1-x^4}\,dx +\int_{0}^{1}\frac{x^2-x}{x^4-1}\,dx = \int_{0}^{1}\frac{dx}{1+x^2}=\frac{\pi}{4}$$ by just breaking the integration range as $(0,1)\cup (1,+\infty)$ and applying the substitution $x\mapsto\frac{1}{x}$ on the second "half".