Confusion Regarding Comparison Theorem

Question

Determine if the integral converges or diverges using the Comparison Theorem (CT).

$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$

I saw solutions online arrive at the following inequality

$$\frac{x+1}{\sqrt{x^4 -x}} \ge \frac{1}{x}$$

But this inequality only holds for $x > 1$ and not for $x \ge 1$ since at $x=1$ the LHS of the inequality, $\frac{x+1}{\sqrt{x^4 -x}}$, is undefined.

I thought the inequality had to be true over the limits of integration, $x \ge 1$ or $[1, \infty]$ in this case, in order to use the CT.

Furthermore, the CT requires the functions being compared to be continuous over a closed interval, but at $x = 1$ function $\frac{x+1}{\sqrt{x^4 -x}}$ is infinitely discontinuous (vetical asymptote) which I'm assuming means we can't apply the CT over $[1, \infty]$.

But I have seen solutions online show that

$$\int_{1}^{\infty} \frac{1}{x}dx$$

diverges and thus $$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$ diverges by the CT. Are these solutions correct?

As a work around I was thinking the inequality is true for $x \ge 2$ so the integral could be "split up" into the following

$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx = \int_{1}^{2} \frac{x+1}{\sqrt{x^4 -x}}dx + \int_{2}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$

The CT could then be applied to the second integral. The first integral, however, still has an infinite discontinuity at $x = 1$ and thus is an "improper" integral. This would require evaluating the integral directly which defeats the purpose of using the CT in the first place to determine convergence or divergence.

My questions are:

1. Does it matter if the inequality is not true over the interval or limits of integration $[a, \infty)$ or $x \ge a$ in order to apply the Comparison Theorem?

2. Were the solutions online correct?

Answer 1

I cannot tell you whether or not the online solutions are correct, since I did not sse them. But I can tell you that your idea of doing the decomposition$$\int_1^\infty \frac{x+1}{\sqrt{x^4 -x}}\,\mathrm dx=\int_1^2\frac{x+1}{\sqrt{x^4 -x}}\,\mathrm dx+\int_2^\infty\frac{x+1}{\sqrt{x^4-x}}\,\mathrm dx$$is fine. Then, you can indeed deduce from the inequality$$\frac{x+1}{\sqrt{x^4-x}}\geqslant\frac1x$$that the second integral diverges. On the other hand$$\lim_{x\to1}\frac{\frac{x+1}{\sqrt{x^4-x}}}{\frac1{\sqrt{1-x}}}=2$$and therefore, since the integral$$\int_1^2\frac{\mathrm dx}{\sqrt{1-x}}$$converges, then so does$$\int_1^2\frac{x+1}{\sqrt{x^4 -x}}\,\mathrm dx.\tag1$$The final conclusion is, of course, that the original integral diverges.

Note that it would still diverge if the integral $(1)$ was divergent. So, in fact, you do not have to analyze the convergence of $(1)$.

Answer 2

Let $\epsilon > 0$ (something very small perhaps) and then

\begin{align} \int_{1}^{n} \frac{x+1}{\sqrt{x^4 -x}}dx & = \int_{1+\epsilon}^{n} \frac{x+1}{\sqrt{x^4 -x}} + \int_{1}^{1 + \epsilon} \frac{x+1}{\sqrt{x^4 -x}} \\ \end{align}

You could then show that (a sentence or two)

$$\int_{1}^{1 + \epsilon} \frac{x+1}{\sqrt{x^4 -x}} \geq 0$$

and that

$$\int_{1+\epsilon}^{n} \frac{x+1}{\sqrt{x^4 -x}} \geq \int_{1 + \epsilon}^n\frac{1}{x}dx $$

diverges (which you have already done). A way to think about it (this certainly isn't rigorous) might be that

$$\frac{x+1}{\sqrt{x^4 - x}} \geq \frac{1}{x}$$

holds apart from at a single point, and given that a single point isn't going to add anything to the integral we can simply ignore it.

Answer 3

Considering the integrand, we have $$\frac{x+1}{\sqrt{x^4 -x}}=\frac{2}{\sqrt{3} \sqrt{x-1}}-\frac{\sqrt{x-1}}{\sqrt{3}}+\frac{2 (x-1)^{3/2}}{3 \sqrt{3}}+O\left((x-1)^{5/2}\right)$$ So, no problem around the lower bound.

On the other side $$\frac{x+1}{\sqrt{x^4 -x}}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{2x^4}+O\left(\frac{1}{x^5}\right)$$ (in fact, in this last expansion, all coefficients are non-negative- so you can bound the integrand by wathever you want).

Asymptotically, we have $$\int_1^n\frac{x+1}{\sqrt{x^4 -x}}=\log(n)+ C$$ where $C\sim 1.864$.