# Question

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

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

# 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.

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

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

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.

• 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? – baxx Jul 21 '17 at 20:18
• Exactly........ – marty cohen Jul 21 '17 at 20:36
• 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 – baxx Jul 21 '17 at 20:38
• You are welcome. – marty cohen Jul 22 '17 at 4:43

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 $n\ne\frac1{13}$

• 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 – baxx Jul 21 '17 at 20:00
• @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. – Simply Beautiful Art Jul 21 '17 at 20:06
• Ok, those haven't been covered yet (they're in the series section). So perhaps I'll understand this answer later on. – baxx Jul 21 '17 at 20:09
• @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 – Simply Beautiful Art Jul 21 '17 at 20:10
• Okay then :-) If you understand any of the other answers, feel free to accept them as well. – Simply Beautiful Art Jul 21 '17 at 20:13

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 $n=1/13$

• I believe you should keep the $n-1/13$ inside the limit. – Simply Beautiful Art Jul 20 '17 at 18:02
• @SimplyBeautifulArt Thank you; I edited accordingly. – Mark Viola Jul 20 '17 at 18:03
• hrm, I don't think that I would have done this algebra :/ – baxx Jul 21 '17 at 19:04
• @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. – Mark Viola Jul 21 '17 at 20:17
• 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. – baxx Jul 21 '17 at 20:30

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