Qs: Use the limit comparison test to determine whether the following integrals converge

$$\int_{x=0}^\infty \frac{x^\frac{1}{3}+5}{x^2-6x+10}$$

I did the question and found that it diverges, but it turns out this integral converges by the Limit Comparison Test.

My Answer: f(x) = (the equation above) and g(x) = $\frac{1}{x^\frac{5}{3}}$

$$\lim_{x\to \infty} f(x)/g(x) = 1$$ which is positive and finite.

$$\int_{x=0}^\infty \frac{1}{x^\frac{5}{3}} diverges$$

So by the LCT, my original integral diverges

However, this integral converges by the LCT and I just don't see how. Can anyone help?

  • $\begingroup$ What is the exact statement of the limit comparison for integral you have in your mind? Because $g(x) = x^{-5/3}$ only tells the asymptotic behavior of your integrand $f(x)$ for large $x$, hence the behavior of $g(x)$ near $x = 0$ is completely irrelevant to that of $f$. $\endgroup$ – Sangchul Lee Mar 28 '18 at 20:32
  • $\begingroup$ Very sorry for my late reply. So, I wanted to show that the integrand $f(x)$ above converges on the range $[0,inf)$. Therefore, I need to have a function $g(x)$ that also converges on this integral. That is why behaviour of $g(x)$ near $x=0$ mattered for me. Does that make sense? $\endgroup$ – mathsexam2013 Apr 3 '18 at 15:41

The trick is that, look at the neighborhood of $x=0$, the integrand is continuous, so just split it: \begin{align*} \int_{0}^{\infty}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx=\int_{0}^{1}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx+\int_{1}^{\infty}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx, \end{align*} and \begin{align*} \int_{0}^{1}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx \end{align*} exists by the continuity of $\dfrac{x^{1/3}+5}{x^{2}-6x+10}$ on the compact set $[0,1]$.

Now use Limit Comparison Test to $1/x^{5/3}$ on $[1,\infty)$ to conclude that \begin{align*} \int_{1}^{\infty}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx \end{align*} exists.

  • $\begingroup$ I understand the 2nd half, about using LCT for [1,inf). However, I am a bit confused as to why the first integral from [0,1] exists. $\endgroup$ – mathsexam2013 Mar 28 '18 at 18:54
  • $\begingroup$ The integrand is continuous, it has no singularity. $\endgroup$ – user284331 Mar 28 '18 at 18:55
  • $\begingroup$ It is simply the usual Riemann integration, not improper Riemann for that. $\endgroup$ – user284331 Mar 28 '18 at 18:55
  • $\begingroup$ In fact, the answers I have did the LCT for [1,inf) then straight away said that the integral exists for [0,inf). I'm guessing that's the extra it missed out, but I have never seen this theorem. Is there an easier way to explain it? $\endgroup$ – mathsexam2013 Mar 28 '18 at 18:56
  • $\begingroup$ Ah right I think I understand now. Would the method of splitting up the integral be best then, for limits of [0,inf)? $\endgroup$ – mathsexam2013 Mar 28 '18 at 18:57

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