Help me to finish calculating $\int_0^{\infty} \frac{1}{x^3-1}dx$

Question

$$\int_0^{\infty} \frac{1}{x^3-1}dx$$

What I did:

$$\lim_{\epsilon\to0}\int_0^{1-\epsilon} \frac{1}{x^3-1}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx$$

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$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx$$

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$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx$$

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$$[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$

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$$[\frac{1}{6}(2ln(x-1)-ln(x^2+x+1))-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln{2(x-1})-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$

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$$[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$

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$$[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$

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$$\lim_{\epsilon\to0}[\frac{1}{6}ln(\frac{(1-\epsilon)^2-2(1-\epsilon)+1}{(1-\epsilon)^2+1-\epsilon+1})-\frac{1}{\sqrt3}\arctan(\frac{2(1-\epsilon)+1}{\sqrt3})+\frac{1}{\sqrt3}\arctan(\frac{1}{\sqrt3})]+\lim_{\epsilon\to 0} [ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})+\cdots]$$

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This is where my problem is, what is :

$$ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})$$

^^^ If I know past this, I know how to proceed. The only thing stopping me is this ^^^. Please help.

Answer 1

We can in fact evaluate the Cauchy principal value of the integral as follows (which is what I think you were trying to do).

Consider the following contour integral:

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

$C$ is a modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$. The modification lies on small semicircular bumps above and below $z=1$ of radius $\epsilon$, and we will consider the limits as $\epsilon \to 0$ and $R\to\infty$.

Let's evaluate this integral over the contours. There are $8$ pieces to evaluate, as follows:

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

(To see this, draw the contour out, including the bumps about $z=1$.)

As $R \to \infty$, the fourth integral vanishes as $\log{R}/R^2$. As $\epsilon \to 0$, the second integral vanishes as it is $O(\epsilon^2)$, while the eighth integral vanishes as $\epsilon \log{\epsilon}$. This leaves the first, third, fifth, sixth and seventh integrals, which in the above limits, become

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

It should be appreciated that, in the fifth, sixth, and seventh integrals, the $i 2 \pi $ factor appears because, on the lower branch of the real axis, we write $z=x \, e^{i 2 \pi}$. In the sixth integral, in fact, $z = e^{i 2 \pi} + \epsilon \, e^{i \phi + 2 \pi}$.

The $PV$ denotes the Cauchy principal value of the integral. As it stands, the integral does not actually converge. Nevertheless, we are not actually considering the integral straight through the pole at $z=1$, but a very small detour around the pole. Thus, in the limit, we get the Cauchy PV. A little rearranging cancels the $\log$ term, and we now have:

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

The contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles. The poles here are at $z=e^{i 2 \pi/3}$ and $z=e^{i 4 \pi/3}$. Note that the pole at $z=1$ is not inside the contour $C$ because of the detour around that "pole". It should be appreciated that the poles must have their arguments between $[0,2 \pi]$ because of the way we defined $C$.

In any case, we now have that the above 1D integrals over the positive real line are equal to

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

We may now solve for the principal value and get:

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

Answer 2

• Subdivide $(0,\infty)$ into $(0,1)$ and $(1,\infty)$.
• On the latter, let $t=\dfrac1x$.
• $a^3-b^3=\big(a-b\big)\Big(a^2+ab+b^2\Big)$.
• Complete the square in the denominator.

Answer 3

Using partial fractions directly I think it is simpler:

$$\int\frac1{x^3-1}dx=\frac13\left(\int\frac1{x-1}-\frac{x+2}{x^2+x+1}\right)dx=$$

$$\frac13\left(\log(x-1)-\frac12\int\frac{2x+1}{x^2+x+1}dx-\frac32\int\frac1{\left(x+\frac12\right)^2+\frac34}dx\right)=$$

$$=\frac13\log(x-1)-\frac12\log(x^2+x+1)-\sqrt3\int\frac{\frac2{\sqrt3}dx}{1+\left(\frac{2x+1}{\sqrt3}\right)^2}=$$

$$=\log\frac{\sqrt[3]{x-1}}{\sqrt{x^2+x+1}}-\arctan\frac{2x+1}{\sqrt3}+C , $$

and etc.

Answer 4

This integral is not defined. You can't write $$\int_0^{\infty} \frac{1}{x^3-1}dx=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon} \frac{1}{x^3-1}dx+\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx\right)$$ (Note that although you initially write two separate limits, you combine them into one limit in a later step, so you are actually working with what I have above. Either that or you correctly started out with two limits, but then later incorrectly combined $$\lim_{\epsilon\to0^+}f(\epsilon)+ \lim_{\epsilon\to0^+}g(\epsilon)=\lim_{\epsilon\to0^+}\left(f(\epsilon)+g(\epsilon)\right)$$ without verifying the two limits each exist.)

This would be true if you have established the first integral exists in the first place, but it does not. Note that this setup has the limiting variables approaching the pole at $x=1$ at the same rate from either side. This is artificially creating cancellation as $x\to1^-$ in the one integral and $x\to1^+$ in the other. Something like $$\int_0^{\infty} \frac{1}{x^3-1}dx=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon^2} \frac{1}{x^3-1}dx+\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx\right)$$ should be just as valid as the first equation, but here the result will be $-\infty$ instead of the finite answers others have found ($0$ and $-{\frac{\pi}{3\sqrt{3}}}$).

Instead, you can write $$\begin{align} \int_0^{\infty} \frac{1}{x^3-1}dx &=\int_0^{1} \frac{1}{x^3-1}dx+\int_1^{\infty} \frac{1}{x^3-1}dx&\text{(provided both exist)}\\ &=\lim_{\epsilon_1\to0^+}\int_0^{1-\epsilon_1} \frac{1}{x^3-1}dx+\lim_{\epsilon_2\to0^+}\int_{1+\epsilon_2}^{\infty} \frac{1}{x^3-1}dx\\ &=\lim_{\epsilon_1\to0^+}\int_0^{1-\epsilon_1} \frac{1}{x^3-1}dx+\lim_{\epsilon_2\to0^+}\lim_{\epsilon_3\to\infty}\int_{1+\epsilon_2}^{\epsilon_3} \frac{1}{x^3-1}dx \end{align}$$ Note that the limiting variables are different. Neither of these improper integrals exist, since the integrands behave like $\frac{c}{x-1}$ near $x=1$.

Answer 5

Cauchy Principal Value

The integral, as written, diverges. In the case of an improper integral such as this, $$ \int_0^\infty\frac1{x^3-1}\mathrm{d}x =\int_0^1\frac1{x^3-1}\mathrm{d}x+\int_1^\infty\frac1{x^3-1}\mathrm{d}x\tag{1} $$ however, neither of the integrals on the right converge.

On the other hand, if what we want is the Cauchy Principal Value, then we are asking for $$ \lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac1{x^3-1}\mathrm{d}x+\int_{1+\epsilon}^\infty\frac1{x^3-1}\mathrm{d}x\right)\tag{2} $$ In many cases, this will exist, even when the actual integral fails to converge.

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A Real Approach to Computing the Cauchy Principal Value

Substituting $x\mapsto\frac1x$, we get $$ \begin{align} \int_{1+\epsilon}^\infty\frac1{x^3-1}\mathrm{d}x &=\int_0^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x\\ &=\int_0^{1-\epsilon}\frac{x}{1-x^3}\mathrm{d}x+\color{#C00000}{\int_{1-\epsilon}^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x}\tag{3} \end{align} $$ Since $\frac1{1+\epsilon}-(1-\epsilon)=\frac{\epsilon^2}{1+\epsilon}$, we have that $$ \begin{align} \color{#C00000}{\int_{1-\epsilon}^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x} &\le\frac{\epsilon^2}{1+\epsilon}\frac{\frac1{1+\epsilon}}{1-\left(\frac1{1+\epsilon}\right)^3}\\ &=\frac{\epsilon(1+\epsilon)}{3+3\epsilon+\epsilon^2}\\[9pt] &\stackrel{\epsilon\to0^+}{\to}0\tag{4} \end{align} $$ Therefore, using $(3)$, $(4)$, and $x+\frac12=\frac{\sqrt3}2\tan(\theta)$, we can conclude $$ \begin{align} &\mathrm{PV}\int_0^\infty\frac1{x^3-1}\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac1{x^3-1}\mathrm{d}x +\int_{1+\epsilon}^\infty\frac1{x^3-1}\mathrm{d}x\right)\\ &=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac1{x^3-1}\mathrm{d}x -\int_0^{1-\epsilon}\frac{x}{x^3-1}\mathrm{d}x +\color{#C00000}{\int_{1-\epsilon}^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x}\right)\\ &=\lim_{\epsilon\to0^+}\left(-\int_0^{1-\epsilon}\frac{x-1}{x^3-1}\mathrm{d}x\right) +\color{#C00000}{0}\\ &=-\int_0^1\frac1{x^2+x+1}\mathrm{d}x\\ &=-\int_0^1\frac1{(x+\frac12)^2+\frac34}\mathrm{d}x\\ &=-\frac2{\sqrt3}\int_{\pi/6}^{\pi/3}1\mathrm{d}\theta\\[9pt] &=-\frac\pi{3\sqrt3}\tag{5} \end{align} $$

Answer 6

Remember that in any limit to infinity only the highest power term in the numerator and denominator matter. Thus your expression is equivalent to $\frac{1}{6}ln(1)=0$

Answer 7

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\mrm{P.V.}\int_{0}^{\infty} {\dd x \over x^{3} - 1}} = \Re\int_{0}^{\infty}{\dd x \over \pars{x - 1 + \ic 0^{+}}\pars{x^{2} + x + 1}} \\[5mm] = &\ -\Re\int_{\infty}^{0}{\ic\,\dd y \over \pars{\ic y - 1 + \ic 0^{+}}\pars{-y^{2} + \ic y + 1}} = \Im\int_{0}^{\infty}{\dd y \over 1 + \ic y^{3}} \\[5mm] = &\ {1 \over 3}\,\Im\int_{0}^{\infty}{y^{\color{red}{1/3} - 1} \over 1 + \ic y}\,\dd y \end{align} Note that $\ds{{1 \over 1 + \ic y} = \sum_{k = 0}^{\infty} \color{red}{\Gamma\pars{1 + k}\expo{\ic\pi k/2}} \,\,{\pars{-y}^{k} \over k!}}$ \begin{align} &\bbox[5px,#ffd]{\mrm{P.V.}\int_{0}^{\infty} {\dd x \over x^{3} - 1}} = {1 \over 3}\,\Im\bracks{% \Gamma\pars{\color{red}{1 \over 3}} \Gamma\pars{1 - \color{red}{1 \over 3}} \expo{\ic\pi\pars{\color{red}{-1/3}}/2}\,} \\[5mm] = &\ {1 \over 3}\bracks{-\sin\pars{\pi \over 6}} {\pi \over \sin\pars{\pi/3}} = {1 \over 3}\pars{-\,{1 \over 2}}\,{\pi \over \root{3}/2} \\[5mm] = &\ \bbx{-\,{\root{3} \over 9}\,\pi} \approx -0.6046 \\ & \end{align}