# Convergence of integral $\int_{0}^{\infty}{\frac{1}{x^{3}-1}}dx$

So i have the integral $$\int_{0}^{\infty}{\frac{1}{x^{3}-1}} dx$$

Software programs say that it is divergent, except for one program which evaluate numerically and gave a clear result.

The integrals $\int_{0}^{1}$ and $\int_{1}^{\infty}$ are indeed divergent if taken separately, but the fact that one goes to infinity and the other to minus infinity, makes me think that the final integral might be convergent.

Also, expanding the two parts in McLaurin series and integrating on appropriate intervals, i find that the whole thing converges to $\frac{\pi\sqrt{3}}{9}$, which is the same result found by the method of contour integration in the complex plane.

So my question is: what is the truth here? Is it convergent or not?

Any idea would be greatly appreciated.

Thanks :)

LATER EDIT:

Thanks to @Ron Gordon,I found a solution for the general case $\int_{0}^{\infty}{\frac{1}{x^{3}-a^{3}}} dx , a>0$, which I will describe here.

Consider the integral:$$\int_{0}^{\infty}{\frac{\ln{z}}{z^3-a^{3}}}dz$$ and the contour:

The green parts of the integral are zero, as showed by @Ron Gordon. So we are left with: $$\int_0^{a-\epsilon}{\frac{\ln z}{z^3-a^3}}dz+ \int_{\pi}^{0}{\frac{\ln(a+\epsilon e^{i\theta})}{(a+\epsilon e^{i\theta})^3-a^3}}i\epsilon e^{i\theta}d\theta+ \int_{a+\epsilon}^{\infty}{\frac{\ln z}{z^3-a^3}}dz$$

$$+\int_\infty^{a+\epsilon}{\frac{\ln z+2\pi i}{z^3-a^3}}dz+ \int_{2\pi}^{\pi}{\frac{\ln(a+\epsilon e^{i\theta})+2\pi i}{(a+\epsilon e^{i\theta})^3-a^3}}i\epsilon e^{i\theta}d\theta+ \int_{a-\epsilon}^0{\frac{\ln z+2\pi i}{z^3-a^3}}dz$$

The sum of middle integrals reduces to (after expanding the denominator, simplifying and keeping in mind that $\epsilon$ tends to $0$) $$\int_\pi^0{\frac{i\ln a}{3a^2}}d\theta+ \int_{2\pi}^{\pi}{\frac{i\ln a-2\pi}{3a^2}}d\theta=-{\frac{2\pi i \ln a}{3a^2}}+{\frac{2\pi^2}{3a^2}}$$

The other four integrals reduce to $$-2\pi i \int_{0}^{a-\epsilon}{\frac{1}{x^3-a^3}}dx-2\pi i\int_{a+\epsilon}^{\infty}{\frac{1}{x^3-a^3}}dx=-2\pi iJ$$

Now, the sum of all these integrals must be equal to the sum of residues in the contour times $2\pi i$, $$-2\pi iJ-{\frac{2\pi i \ln a}{3a^2}}+{\frac{2\pi^2}{3a^2}}=2\pi i\sum{Res}$$

We have two residues to find, at the points $a e^{i{\frac{2\pi}{3}}}$ and $a e^{i{\frac{4\pi}{3}}}$. Their sum is $$\sum{Res}={\frac{\ln (ae^{i2\pi/3})}{3a^2e^{i2\pi/3}}}+{\frac{\ln (ae^{i4\pi/3})}{3a^2e^{i4\pi/3}}}={\frac{{\frac{\pi \sqrt{3}}{3}}-\ln a-\pi i}{3a^2}}$$

Multiplying this by $2\pi i$ and equating with the sum of integrals, we find that $$J=-\frac{\pi\sqrt{3}}{9a^2}$$

So the integral is convergent after all (in the sense of principal value of course). Thanks everyone for the help.

• It is not convergent. However you can calculate its principal value. en.wikipedia.org/wiki/Cauchy_principal_value – Pocho la pantera Sep 10 '13 at 15:35
• The Maple command $$int(1/(x^3-1), x = 0 .. infinity, CauchyPrincipalValue) ;$$ gives $-1/9\,\pi \,\sqrt {3}$. – user64494 Sep 10 '13 at 15:42
• negative infinity + positive infinity =/= 0 – Don Larynx Sep 11 '13 at 6:48

This integral is improper in two different ways: you have a problem at $x=1$, and the fact that your region of integration is unbounded.

The fact that the region is unbounded doesn't end up being a problem; but, consider $$\int_1^2\frac{1}{x^3-1}\,dx.$$ Note that you can write $x^3-1=(x-1)(x^2+x+1)$. For $x\in[1,2]$, $x^2+x+1\leq 2^2+2+1=7$, so that $$\frac{1}{x^3-1}\geq\frac{1}{7(x-1)}\geq0,\qquad 1<x\leq 2.$$ Can you see why $\int_1^2\frac{1}{x-1}\,dx$ diverges to $\infty$? That, the inequality, and non-negativity combine to demand that $\int_1^2\frac{1}{x^3-1}\,dx$ does the same.

The program returned the Cauchy Principal Value. Other than that it is an improper integral.

• Ramanujan master theorem should give $PV \int_{0}^{\infty} \frac{x^{s-1}}{x-a}= a^{s-1}\pi cotg( \pi s)$ – Jose Garcia Sep 10 '13 at 21:48

The integral diverges due to the pole at $x=1$. You have, however, the wrong sign (and thus value) for the Cauchy principal value, as I will demonstrate.

Consider the integral in the complex plane

$$\oint_C dz \frac{\log{z}}{z^3-1}$$

where $C$ is a keyhole contour in the complex plane which has a small semicircular divot under the real axis at $z=1$. The contour $C$ then consists of six pieces:

$$\int_{\epsilon}^R dx \frac{\log{x}}{x^3-1} + i R \int_0^{2 \pi} d\theta\, e^{i \theta} \frac{\log{R}+i \theta}{R^3 e^{i 3 \theta}-1} + \\ \int_R^{1+\epsilon} dx \frac{\log{x}+i 2 \pi}{x^3-1} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\log{(1+\epsilon e^{i \phi})}+i 2 \pi}{(1+\epsilon e^{i \phi})^3-1} + \\ \int_{1-\epsilon}^{\epsilon}dx \frac{\log{x}+i 2 \pi}{x^3-1} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log{\epsilon}+i \phi}{\epsilon^3 e^{i 3 \phi}-1}$$

We seek the limits as $R \to \infty$ and $\epsilon \to 0$. Note that $\log{x}=0$ at $x=1$, so we need no principal value above the real axis. Below, however, the $i 2 \pi$ piece does not remove the singularity, so we must use the divot there; the divot is represented by the fourth integral.

It should be clear that the second integral vanishes as $\log{R}/R^2$ and may be neglected. The sixth integral vanishes as $\epsilon \log{\epsilon}$ and may also be neglected. The first, third, and fifth integral combine to produce $-i 2 \pi$ times our Cauchy principal value. The fourth integral, that around the divot, becomes

$$i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\epsilon e^{i \phi} + i 2 \pi}{3 \epsilon e^{i \phi}} \sim \frac{2 \pi^2}{3} \quad (\epsilon \to 0)$$

Thus, the contour integral takes the value

$$-i 2 \pi PV \int_0^{\infty} \frac{dx}{x^3-1} + \frac{2 \pi^2}{3}$$

This is equal to $i 2 \pi$ times the sum of the residues at the remaining poles $z=e^{i 2 \pi/3}$ and $z=e^{i 4 \pi/3}$, which is

$$i 2 \pi \left (\frac{i 2 \pi/3}{3 e^{i 4 \pi/3}} + \frac{i 4 \pi/3}{3 e^{i 8 \pi/3}} \right )= \frac{2 \pi^3}{3}+i \frac{2 \pi^2}{3 \sqrt{3}}$$

We therefore have

$$PV \int_0^{\infty} \frac{dx}{x^3-1}=-\frac{\pi}{3 \sqrt{3}}$$

Note the minus sign in the answer has been verified with WA.

• Hi thanks for your answer. You said 'Note that logx=0 at x=1, so we need no principal value above the real axis'. I dont understand why we need no divot above the axis, because I just did the integral for a general case, and it gave the wrong answer without the divot above. I will try to edit my question to include the solution I found – gun Sep 10 '13 at 19:29
• @gun: as I stated, no divot needed there because of the log term. But due to the multivaluedness exploited, the $i 2 \pi$ term requires the divot under the real axis. – Ron Gordon Sep 10 '13 at 19:40
• Ok, i think i understand what you are saying, but what if it was an arbitrary number $a>0$, like in $\frac{1}{x^3-a^3}$? The solution I showed in the edited question wouldnt have worked if i would not make a divot above. Or maybe i did something wrong? – gun Sep 10 '13 at 21:48
• @gun: you are right. In the general case, the log does not remove the singularity, so that is the contour I would use as well. Good work. – Ron Gordon Sep 10 '13 at 23:52

The integral is absolutely divergent: \begin{align} \int_0^\infty\frac1{|x^3-1|}\,\mathrm{d}x &\ge\int_{1+1/n}^2\frac1{|x^3-1|}\,\mathrm{d}x\\ &=\int_{1+1/n}^2\frac1{(x-1)(x^2+x+1)}\,\mathrm{d}x\\ &\ge\frac17\int_{1+1/n}^2\frac1{x-1}\,\mathrm{d}x\\ &=\frac17\log(n)\tag{1} \end{align} Inequality $(1)$ is true for any $n\ge1$. Therefore, the integral of the absolute value diverges.

However, just as a series can diverge absolutely yet converge conditionally, we can apply what is termed the Cauchy Principal Value. That is, we take advantage of the cancellation near the singularity at $x=1$ by cutting out a symmetric region around $1$: \begin{align} \hspace{-1cm}\mathrm{PV}\int_0^\infty\frac1{x^3-1}\,\mathrm{d}x &=\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac1{x^3-1}\,\mathrm{d}x +\int_{1+\epsilon}^\infty\frac1{x^3-1}\,\mathrm{d}x\\ &=\lim_{\epsilon\to0}\int_{1+\epsilon}^2\left(\frac1{x^3-1}+\frac1{(2-x)^3-1}\right)\,\mathrm{d}x +\int_2^\infty\frac1{x^3-1}\,\mathrm{d}x\\ &=\int_1^2\frac{-6}{(7-5x+x^2)(1+x+x^2)}\,\mathrm{d}x +\int_2^\infty\frac1{x^3-1}\,\mathrm{d}x\tag{2} \end{align} Neither integral in the last line of $(2)$ has a singularity in the domain of integration, and the denominators are greater than order $1$, so they converge.

For comparison, Mathematica says that the integral in $(2)$ is $$-\frac\pi{3\sqrt3}\tag{3}$$

\begin{align} {\cal P}\int_{0}^{\infty}{{\rm d}x \over x^{3} - 1} &= \Re\int_{0}^{\infty}{{\rm d}x \over x^{3} - 1 + {\rm i}0^{+}} = \Re\left(\int_{0}^{1}{{\rm d}x \over x^{3} - 1 + {\rm i}0^{+}} + \int_{0}^{1} {\left(1/x^{2}\right)\,{\rm d}x \over \left(1/x^{3}\right) - 1 + {\rm i}0^{+}}\right) \\[3mm]&= {\cal P}\int_{0}^{1}{1 - x \over x^{3} - 1}\,{\rm d}x = -\int_{0}^{1}{{\rm d}x \over x^{2} + x + 1} = -\int_{0}^{1}{{\rm d}x \over \left(x + 1/2\right)^{2} + 3/4} \\[3mm]&= -\int_{1/2}^{3/2}{{\rm d}x \over x^{2} + 3/4} = -\,{2\sqrt{3\,} \over 3}\int_{\sqrt{3\,}/3}^{\sqrt{3\,}} {{\rm d}x \over x^{2} + 1} \\[3mm]&= -\,{2\sqrt{3\,} \over 3}\left\lbrack \arctan\left(\sqrt{3\,}\right) - \arctan\left(\sqrt{3\,} \over 3\right) \right\rbrack = -\,{2\sqrt{3\,} \over 3}\left({\pi \over 3} - {\pi \over 6}\right) \\[3mm]&= \color{#ff0000}{\large% -\,{\sqrt{3\,} \over 9}\, = -\,{\pi \over 3\sqrt{3\,}}} \end{align}

${\bf\mbox{This is the}\ {\large\underline{\mbox{Mathematica}}}\ \mbox{result}}$.