Showing $\int_{-\infty}^{\infty}\frac {\mathrm d x} {1+x+x^2}=3 \int_0^\infty \frac {\mathrm d x} {1+x+x^2}$

Question

Is there an easy way to see why $$\int_{-\infty}^{\infty}\frac {\mathrm d x} {1+x+x^2}=3 \int_0^\infty \frac {\mathrm d x} {1+x+x^2}$$ without having to evaluate the integrals explicitly? I was trying substitutions but nothing worked.

Answer 1

The key is the equality:

$$(x^2+x+1)(x^2-x+1)=x^4+x^2+1$$

Letting $$\begin{align}I&=\int_{0}^{\infty}\frac{dx}{1+x+x^2}\\J&=\int_{-\infty}^{0}\frac{dx}{1+x+x^2}\\&=\int_{0}^{\infty}\frac{dx}{1-x+x^2}\end{align}$$

Then you want to show that $I+J=3I.$ We'll show the equivalent $J-I=I.$

We can see that:

$$J-I=\int_{0}^{\infty}\frac{2x}{1+x^2+x^4}\,dx$$

Substituting $u=x^2$ you get:

$$J-I=\int_{0}^{\infty} \frac{du}{1+u+u^2}=I$$ or $J=2I.$

Answer 2

Let $\displaystyle x=\frac{1}{v}$.

$$\int_{-1}^0\frac{dx}{1+x+x^2}=\int_{-1}^{-\infty}\frac{\frac{-1}{v^2}dv}{1+\frac{1}{v}+\frac{1}{v^2}}=\int_{-\infty}^{-1}\frac{dv}{1+v+v^2}$$

Let $\displaystyle x=-1-u$.

$$\int_{-\infty}^{-1}\frac{dx}{1+x+x^2}=\int_\infty^0\frac{-du}{1+(-1-u)+(-1-u)^2}=\int_0^{\infty}\frac{du}{1+u+u^2}$$

Therefore,

\begin{align*} \int_{-\infty}^{\infty}\frac{dx}{1+x+x^2}&=\int_{-\infty}^{-1}\frac{dx}{1+x+x^2}+\int_{-1}^{0}\frac{dx}{1+x+x^2}+\int_{0}^{\infty}\frac{dx}{1+x+x^2}\\ &=3\int_{0}^{\infty}\frac{dx}{1+x+x^2} \end{align*}