Proof for for convergence of $\int_{-∞}^0\frac{\sqrt{\left|x\right|}}{x^2+x+1}dx$

Question

I want to show that the following integral converges. However I am troubled finding an estimate for this term which is sufficient for the proof. Could I use L'Hospital on this one?

$$\int_{-∞}^0\frac{\sqrt{\left|x\right|}}{x^2+x+1}dx$$

Answer 1

$$ \underbrace{x^2+x+1 = \left( x + \frac 1 2 \right)^2 + \frac 3 4}_\text{completing the square} = (\text{a square}) + (\text{a positive number}) $$ The denominator is nowhere $0,$ so you have no vertical asymptotes. Thus the function is continuous on $[a,0]$ if $a$ is any negative number. Thus we only need to examine what happens on $(-\infty,a].$

$$ \frac{\sqrt{-x}}{\left( x + \frac 1 2 \right)^2 + \frac 3 4} \le \frac 1 {\left(x + \frac 1 2 \right)^{3/2}} \text{ and } \int_{-\infty}^a \frac{dx}{\left(x+ \frac 1 2 \right)^{3/2}} \text{ converges.} $$

Answer 2

$$\int_{-\infty}^{0}\frac{\sqrt{|x|}\,dx}{x^2+x+1} = \int_{0}^{+\infty}\frac{\sqrt{x}\,dx}{x^2-x+1} \stackrel{x\mapsto z^2}{=} \int_{0}^{+\infty}\frac{2z^2}{z^4-z^2+1}\,dz$$ equals (by parity) $$\int_{-\infty}^{+\infty}\frac{dz}{z^2+\frac{1}{z^2}-1}=\int_{\mathbb{R}}\frac{dz}{\left(z-\frac{1}{z}\right)^2+1}\stackrel{\text{GMT}}{=}\int_{\mathbb{R}}\frac{dz}{z^2+1}=\color{red}{\pi}<+\infty. $$ $\text{GMT}$ stands for Glasser's Master Theorem.

Answer 3

Note that $x^2+x+1=\frac{1}{2}(x^2+1)+\frac{1}{2}(x+1)^2\geq\frac{1}{2}(x^2+1)$, and thus $$\int_{-\infty}^0\frac{\sqrt{|x|}}{x^2+x+1}\leq 2\int_{-\infty}^0\frac{\sqrt{|x|}}{x^2+1}.$$ Now, the integral from $-1$ to $0$ converges, and for $|x|>1$ we have $\frac{\sqrt{|x|}}{x^2+1}<\frac{1}{|x|^\frac{3}{2}}$. And we know that the integral $$\int_{-\infty}^{-1}\frac{1}{|x|^\frac{3}{2}}=\Big[\frac{1}{2}\frac{1}{\sqrt{|x|}}\Big]_{-\infty}^{-1}=\frac{1}{2}<\infty$$ converges.

However, I don't understand what you mean by L'Hospital in this connection.