5
$\begingroup$

I'm going through Nahin's book Inside Interesting Integrals, and I'm stuck at an early problem, Challenge Problem 1.2: to show that $$\int_1^\infty \frac{dx}{\sqrt{x^3-1}}$$ exists because there is a finite upper-bound on its value. In particular, show that the integral is less than 4.

I've tried various substitutions and also comparisons with similar integrals, but the problem is that any other integral that I can easily compare the above integral to is just as hard to integrate, which doesn't help solve the problem.

I also tried just looking at the graph and hoping for insight, but that didn't work either.

So how doesone one place an upper bound on the integral?

$\endgroup$
2
  • $\begingroup$ You may also improve the upper bound to $3$. $\endgroup$ Jan 29, 2016 at 9:40
  • 1
    $\begingroup$ Letting $x=\dfrac1t$ we are able to express the result in terms of the beta function. $\endgroup$
    – Lucian
    Jan 29, 2016 at 13:51

4 Answers 4

18
$\begingroup$

$$ \frac{1}{\sqrt{x^3-1}} = \frac{1}{\sqrt{x-1}\sqrt{x^2+x+1}} < \frac{1}{x\sqrt{x-1}}$$

$$ \int_1^\infty \frac{\mathrm{d}x}{x\sqrt{x-1}} = \pi < 4$$

This integral can be done by letting $u=\sqrt{x-1}$ which yields $\mathrm{d}x=2u\mathrm{d}u$,

$$ \int_0^\infty \frac{2 \mathrm{d}u}{u^2+1} = \pi$$

$\endgroup$
2
  • 1
    $\begingroup$ I like this answer. It's nice and simple and gives a better upper bound than 4. Thanks! $\endgroup$
    – feralin
    Jan 29, 2016 at 7:22
  • $\begingroup$ No problem, I enjoyed figuring it out. I love that book you're reading :) $\endgroup$
    – Spencer
    Jan 29, 2016 at 7:23
12
$\begingroup$

$$\begin{eqnarray*}\color{red}{I}=\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-1}}=\int_{0}^{+\infty}\frac{dx}{\sqrt{x^3+3x^2+3x}}&=&\int_{0}^{+\infty}\frac{2\, dz}{\sqrt{z^4+3z^2+3}}\\&\color{red}{\leq}&\int_{0}^{+\infty}\frac{2\,dz}{\sqrt{z^4+3z^2+\frac{9}{4}}}=\color{red}{\pi\sqrt{\frac{2}{3}}.}\end{eqnarray*}$$

A tighter bound follows from Cauchy-Schwarz:

$$\begin{eqnarray*} \color{red}{I} &=&2\int_{0}^{+\infty}\frac{\sqrt{z^2+\sqrt{3}}}{\sqrt{z^4+3z^2+3}}\cdot\frac{dz}{\sqrt{z^2+\sqrt{3}}}\\&\color{red}{\leq}& 2\sqrt{\left(\int_{0}^{+\infty}\frac{z^2+\sqrt{3}}{z^4+3z^2+3}\,dz\right)\cdot\int_{0}^{+\infty}\frac{dz}{z^2+\sqrt{3}}}\\&=&2\pi\cdot\left(\frac{1}{6}-\frac{1}{4\sqrt{3}}\right)^{1/4}\leq\color{red}{\frac{2\pi}{42^{1/4}}}.\end{eqnarray*} $$

The manipulations in the first line show that $I$ is just twice a complete elliptic integral of the first kind, whose value can be computed through the arithmetic-geometric mean.

On the other hand, through the substitution $x=\frac{1}{t}$ and Euler's beta function we have:

$$ I \color{red}{=} \frac{\Gamma\left(\frac{1}{3}\right)\cdot \Gamma\left(\frac{1}{6}\right)}{2\sqrt{3\pi}}\color{red}{\leq}\frac{3\cdot 6}{2\sqrt{3\pi}}=\sqrt{\frac{27}{\pi}}$$

since in a right neighbourhood of the origin we have $\Gamma(x)\leq\frac{1}{x}$.

As a by-product we get:

$$ \frac{\Gamma\left(\frac{1}{3}\right)\cdot \Gamma\left(\frac{1}{6}\right)}{2\sqrt{3\pi}} = \frac{\pi}{\text{AGM}(\frac{1}{2} \sqrt{3+2 \sqrt{3}},3^{1/4})}$$

that allows us to compute $\Gamma\left(\frac{1}{6}\right)$ through an AGM-mean:

$$ \Gamma\left(\frac{1}{6}\right) = \color{red}{\frac{2^{\frac{14}{9}}\cdot 3^{\frac{1}{3}}\cdot \pi^{\frac{5}{6}} }{\text{AGM}\left(1+\sqrt{3},\sqrt{8}\right)^{\frac{2}{3}}}}.$$

The last identity was missing in the Wikipedia page about particular values of the $\Gamma$ function, so I took the liberty to add it and add this answer as a reference.

$\endgroup$
10
  • $\begingroup$ I can see that $\int_0^\infty\frac{1}{x^4+3x^2+3}\;dx$ is an elliptic integral but I am curious as to how you evaluated it. The methods I know for putting a quartic like that into standard form are complicated enough that I can't get to your closed form (although I can see that it's right) working by hand (or even with Wolfram Alpha helping me). Is there a simple rule I'm missing for transforming $x^4+3x^2+3$? $\endgroup$
    – Anon
    Jan 21, 2017 at 2:23
  • $\begingroup$ @Anon: what you wrote is not an elliptic integral, we may evaluate it through residues. However, $\sqrt{z^4+3z^2+3}=\sqrt{(z^2+A)(z^2+B)}$ and $$\int_{0}^{+\infty}\frac{dz}{\sqrt{(z^2+A)(z^2+B)}}$$ is a complete elliptic integral of the first kind, that can be evaluated through the AGM. $\endgroup$ Jan 21, 2017 at 12:40
  • $\begingroup$ Oh yes, a typing error on my part to miss the square root. I was aware of what you wrote; sorry for not being clear. Specifically what I was asking was how you computed the values of $A$ and $B$. I got complex values, and tried to use the methods I knew of transforming into a real-valued expression by hand, but failed. How did you compute it? $\endgroup$
    – Anon
    Jan 22, 2017 at 23:02
  • 1
    $\begingroup$ @Anon: the usual way, $AGM(a,b)=AGM(\sqrt{ab},(a+b)/2)$. $\endgroup$ Jan 23, 2017 at 0:10
  • 1
    $\begingroup$ As a matter of fact the identity follows directly from the expression for $K(k_3)$ since $\frac{1+\sqrt{3}}{\sqrt{8}}=k_3'$ $\endgroup$
    – Anon
    Jan 26, 2017 at 2:16
3
$\begingroup$

We could go for crude. Split into the integral from $1$ to $2$, and the integral from $2$ to $\infty$.

On the interval $2$ to $\infty$ we have $x^3-1\gt x^3/4$, and therefore the easily computed $\int_2^\infty \frac{2}{x^{3/2}}\,dx$ provides an upper bound for that region.

On the interval $1$ to $2$, we have $x^3-1=(x-1)(x^2+x+1)\ge 3(x-1)$. And the integral $\int_1^2 \frac{1}{\sqrt{3(x-1)}}\,dx$ can be computed explicitly.

$\endgroup$
0
$\begingroup$

$\int_1^\infty \frac{dx}{\sqrt{x^3-1}}=\int_1^2 + \int_2^\infty=I_1+I_2.$

$I_2 = \int_2^\infty \frac{dx}{x^{3/2}\sqrt{1-1/x^3}}\color{red}{\leq}\lim_{A \to \infty} \int_2^A \frac{dx}{x^{3/2}\sqrt{1-1/A^3}}=\lim_{A \to \infty}\sqrt{\frac{A^3}{A^3-1}}\int_2^A x^{-3/2}dx$

$I_1=\int_1^2\frac{dx}{(x-1)^{1/2}\sqrt{x^2+x+1}}\le \frac{1}{\sqrt{3}}\int_1^2(x-1)^{-1/2}d(x-1)$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .