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Question

Determine $n$ such that the following improper integral is convergent

$$ \int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} -\frac{1}{13x + 1} \right) dx $$


I'm not sure how to go about this.


Working

This is convergent if

$$ \lim_{b \to + \infty} \int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx $$

exists.

The indefinite integral is

\begin{equation*} \begin{aligned} \int \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right)dx & = \int \left(\frac{nx^2}{x^3 + 1} \right) - \int \left(\frac{1}{13x + 1} \right)dx \\ &= \frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1) \end{aligned} \end{equation*}

Which gives

\begin{equation*} \begin{aligned} &\lim_{b \to + \infty} \int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx \\ &= \lim_{b \to + \infty} \left[\frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1) \right]_1^b \\ &= \lim_{b \to + \infty} \left(\left[\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) \right] - \left[\frac{n}{3} \cdot \ln(2) - \frac{1}{13} \cdot \ln(14) \right] \right) \\ &= \lim_{b \to + \infty} \left(\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) - \frac{n}{3} \cdot \ln(2) + \frac{1}{13} \cdot \ln(14) \right) \\ &= \lim_{b \to + \infty} \left( \frac{n}{3} \left( \ln(b^3 + 1) - \ln(2) \right) - \frac{1}{13} \left( \ln(13b + 1) - \ln(14) \right) \right) \\ &= \lim_{b \to + \infty} \left( \frac{n}{3} \left( \ln \left(\frac{b^3 + 1}{2}\right) \right) - \frac{1}{13} \left( \ln \left(\frac{13b + 1}{14}\right) \right) \right) \end{aligned} \end{equation*}

I've tried to use L'Hopital's from here as I have the form $(+ \infty ) - ( + \infty)$. But things went pretty south.

So I'm sure there's a better approach.

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4 Answers 4

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A simpler approach:

Since we know that:

$$\int_1^\infty\frac1{x^a}~\mathrm dx<\infty\iff a>1$$

It follows that if we know

$$\lim_{x\to\infty}\frac{\frac{nx^2}{x^3+1}-\frac1{13x+1}}{1/x^a}=c\ne0$$

Then the integral converges iff $a>1$.


For $n=\frac1{13}$, we find that using $a=2$ satisfies the limit, so it converges for $n=\frac1{13}$.

For $n\ne\frac1{13}$, we find that using $a=1$ satisfies the limit, so it diverges for $n\ne\frac1{13}$

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  • $\begingroup$ I'm not actually following your reasoning. We know that 1/x^a converges if a > 1, yes. But I'm not really 'joining all the dots up' here. If a > 1 then 1/x^a converges, but what does that say about dividing by 1/x^a , ? Just that I'm dividing by a definite value? How does that tell me anything about the numerator there? Then you go onto "for n = ...", but I've not got the reasoning for the first bit, so I'm not sure where the second part is coming from either. Cheers $\endgroup$
    – baxx
    Jul 21, 2017 at 20:00
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    $\begingroup$ @baxx Its basically the limit comparison test. For large enough $x$, we know that $$\left|\frac{nx^2}{n^3+1}-\frac1{13x+1}-\frac1{x^a}-c\right|<\varepsilon$$And so the integrals are comparable to one another. $\endgroup$ Jul 21, 2017 at 20:06
  • $\begingroup$ Ok, those haven't been covered yet (they're in the series section). So perhaps I'll understand this answer later on. $\endgroup$
    – baxx
    Jul 21, 2017 at 20:09
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    $\begingroup$ @baxx The basic idea is that if the limit exists and is non-zero, then that means the two functions are approximately the same, and so are their integrals $\endgroup$ Jul 21, 2017 at 20:10
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    $\begingroup$ Okay then :-) If you understand any of the other answers, feel free to accept them as well. $\endgroup$ Jul 21, 2017 at 20:13
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Note that we have

$$\begin{align} \lim_{b\to \infty}\left(\frac n3\log\left(b^3+1\right)-\frac1{13}\log\left(13b+1\right)\right) &= -\frac1{13}\log(13)\\\\ &+\lim_{b\to \infty}\left(n\log(b)-\frac1{13}\log(b)\right)\\\\ &+\frac n3 \lim_{b\to \infty}\log\left(1+\frac1{b^3}\right)\\\\&-\lim_{b\to \infty}\log\left(1+\frac{1}{13b}\right)\\\\ &=-\frac1{13}\log(13)+\lim_{b\to \infty}\left((n-1/13)\log(b)\right) \end{align}$$

which converges if and only if $n=1/13$

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    $\begingroup$ I believe you should keep the $n-1/13$ inside the limit. $\endgroup$ Jul 20, 2017 at 18:02
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    $\begingroup$ @SimplyBeautifulArt Thank you; I edited accordingly. $\endgroup$
    – Mark Viola
    Jul 20, 2017 at 18:03
  • $\begingroup$ hrm, I don't think that I would have done this algebra :/ $\endgroup$
    – baxx
    Jul 21, 2017 at 19:04
  • $\begingroup$ @baxx The analysis uses only elementary properties of the logarithm. Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Jul 21, 2017 at 20:17
  • $\begingroup$ I find it hard to follow because of the algebra used, many steps have been taken in one, but I don't want to turn it into an algebra lesson, hopefully it's informative to others. $\endgroup$
    – baxx
    Jul 21, 2017 at 20:30
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$\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} -\frac{1}{13x + 1} \right) dx $

$\frac{nx^2}{x^3 + 1} -\frac{1}{13x + 1} =\frac{nx^2(13x+1)-(x^3 + 1)}{(x^3 + 1)(13x + 1)} =\frac{x^3(13n-1)+nx^2-1}{(x^3 + 1)(13x + 1)} $.

If $13n-1 \ne 0$, this behaves like $\frac1{x}$ and the integral diverges.

Therefore the only $n$ for which the integral might converge is $n =\frac1{13}$. For this $n$, the integrand behaves like $\frac1{x^2}$ and the integral of this does converge.

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  • $\begingroup$ Is this because, if $n \neq (1/13)$ then as $x \to \infty$ the highest power over the lowest power is $x^3 / x^4$ or $1/x$, which is known to be divergent? $\endgroup$
    – baxx
    Jul 21, 2017 at 20:18
  • $\begingroup$ Exactly........ $\endgroup$ Jul 21, 2017 at 20:36
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    $\begingroup$ OK - so the approach here was, rather than evaluating the integral after performing integration, instead to look at the form to be integrated and judge from that. This is much easier to follow, thanks $\endgroup$
    – baxx
    Jul 21, 2017 at 20:38
  • $\begingroup$ You are welcome. $\endgroup$ Jul 22, 2017 at 4:43
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Hint: note $\ln (b^3+1) \sim \ln b^3=3\ln b$ and $\ln (13b+1) \sim \ln 13b=\ln 13+\ln b$ for $b\to+\infty$.

Hence: $$\lim_\limits{b\to +\infty} [\frac{n}{3} \ln (b^3+1)-\frac{1}{13}\ln (13b+1)]=\lim_\limits{b\to+\infty} [\left(n-\frac{1}{13}\right) \ln b - \frac{\ln 13}{13}]=\begin{cases} -\frac{\ln 13}{13}, \ if \ n=\frac{1}{13} \\ -\infty, \ if \ n<\frac{1}{13} \\ +\infty, \ if \ n>\frac{1}{13}\end{cases}.$$

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