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

What I did:

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


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


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


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


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


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


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


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


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.

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  • $\begingroup$ I've no idea what you did in the second line after "What I did". We have that : $$\frac1{x^3-1}=\frac13\left(\frac1{x-1}-\frac{x+2}{x^2+x+1}\right)$$ so where do all those summands come from? $\endgroup$
    – Timbuc
    Nov 12, 2014 at 15:59
  • $\begingroup$ He factored $x^3-1$ and then used partial fractions. $\endgroup$
    – Ari
    Nov 12, 2014 at 16:01
  • $\begingroup$ And that's what I did, @Ari, yet I only get two summands, so either he or I are wrong, or we both are...or we both are right and he did something else I haven't considered. $\endgroup$
    – Timbuc
    Nov 12, 2014 at 16:02
  • $\begingroup$ Just calculating what he has there, he seems to be correct. $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ $\endgroup$
    – Ari
    Nov 12, 2014 at 16:06
  • $\begingroup$ @Ari, can you see that's exactly what I got above, and then I did partial fractions?? Perhaps his partial fractions are the same as mine but ordered in another way... $\endgroup$
    – Timbuc
    Nov 12, 2014 at 16:07

7 Answers 7

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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:

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

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  • $\begingroup$ I would usually go with a contour integration to compute this, but this gave me an opportunity to try a real approach. (+1) $\endgroup$
    – robjohn
    Nov 16, 2014 at 9:03
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  • 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.
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  • $\begingroup$ I have done these already :) $\endgroup$
    – The Artist
    Nov 13, 2014 at 5:12
  • $\begingroup$ @TheArtist: If you would “have done these already”, then you wouldn't have written all those absurd monstrosities which take up several screens in the body of your post, but rather, in less than five steps, you would've arrived at the simple conclusion that $I=-\displaystyle\int_0^1\dfrac{dx}{x^2+x+1}$, whose evaluation is trivial. $\endgroup$
    – Lucian
    Nov 13, 2014 at 6:03
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    $\begingroup$ @TheArtist: In general, for $n>1$ we have $~\displaystyle\int_0^\infty\dfrac{dx}{1-x^n}=\dfrac\pi n\cot\dfrac\pi n$ $\endgroup$
    – Lucian
    Nov 13, 2014 at 7:14
  • $\begingroup$ Ohh , I didn't know that :) Learnt something new, Thank you Lucian :) $\endgroup$
    – The Artist
    Nov 13, 2014 at 8:07
  • $\begingroup$ @Lucian: you mean the Cauchy PV of the integral, right? The integral you wrote diverges. $\endgroup$
    – Ron Gordon
    Nov 13, 2014 at 13:56
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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.

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  • $\begingroup$ Do you see that the second step is exactly my partial fraction? $\endgroup$
    – The Artist
    Nov 12, 2014 at 16:35
  • $\begingroup$ Yup...but it sure looks different. :) $\endgroup$
    – Timbuc
    Nov 12, 2014 at 16:39
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    $\begingroup$ What about the singularity at $x=1$? The integral diverges near $x=1$. $\endgroup$
    – robjohn
    Nov 12, 2014 at 18:50
  • $\begingroup$ @robjohn, I didn't get into the improper integral's calculation, as I was trying to make a point in partial fractions, but yes: it seems the integral's divergent there. This makes the PV calculation more important. $\endgroup$
    – Timbuc
    Nov 12, 2014 at 19:09
  • $\begingroup$ Im sorry, I accidentally downvoted your answer. Now I want to change that, can you can edit your answer (coz my vote is locked), so that I can reverse my vote. Please accept my apology. Thanks :) $\endgroup$
    – The Artist
    Nov 16, 2014 at 2:52
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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 $x=1$.

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  • $\begingroup$ (+1) Thank you very much for pointing it out. I should have stated that Im taking Cauchy Principal Value. $\endgroup$
    – The Artist
    Nov 16, 2014 at 7:15
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    $\begingroup$ I disagree. Of course we may divide the integral into two terms and take the limit as The Artist has done to avoid singularity at $x=1$. I agree that as separated integrals each integrals don't converge but as a single integral it does exist & converge. $\endgroup$
    – Venus
    Nov 16, 2014 at 7:27
  • $\begingroup$ Sorry @Venus, but if you agree that the separate parts don't converge, then you agree that the whole integral is not defined. You are thinking of the Cauchy Principle Value (as OP has clarified that (s)he is after), which is a substitute for getting something out of an undefined integral like this. $\endgroup$
    – 2'5 9'2
    Nov 16, 2014 at 7:32
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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.


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 $$ \begin{align} &\mathrm{PV}\int_0^\infty\frac1{x^3-1}\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac1{x^3-1}\mathrm{d}x +\int_{1+\epsilon}^\infty\frac1{x^3-1}\mathrm{d}x\right)\\ &=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac1{x^3-1}\mathrm{d}x -\int_0^{1-\epsilon}\frac{x}{x^3-1}\mathrm{d}x +\color{#C00000}{\int_{1-\epsilon}^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x}\right)\\ &=\lim_{\epsilon\to0^+}\left(-\int_0^{1-\epsilon}\frac{x-1}{x^3-1}\mathrm{d}x\right) +\color{#C00000}{0}\\ &=-\int_0^1\frac1{x^2+x+1}\mathrm{d}x\\ &=-\int_0^1\frac1{(x+\frac12)^2+\frac34}\mathrm{d}x\\ &=-\frac2{\sqrt3}\int_{\pi/6}^{\pi/3}1\mathrm{d}\theta\\[9pt] &=-\frac\pi{3\sqrt3}\tag{5} \end{align} $$

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  • $\begingroup$ (+1) Thank you Rob John xD this is such a nice answer :) $\endgroup$
    – The Artist
    Nov 16, 2014 at 9:18
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Remember that in any limit to infinity only the highest power term in the numerator and denominator matter. Thus your expression is equivalent to $\frac{1}{6}ln(1)=0$

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  • $\begingroup$ U mean to say $\frac{(\infty)^2}{(\infty)^2}=1$??? $\endgroup$
    – The Artist
    Nov 12, 2014 at 15:59
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    $\begingroup$ Absolutely. $\frac{x^2}{x^2}=1$ for all $x$, even as they approach infinity. $\endgroup$
    – Ari
    Nov 12, 2014 at 16:00
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    $\begingroup$ It might make more sense to write $\frac{x^2-2x+1}{x^2+x+1}=\frac{1-2/x+1/x^2}{1+1/x+1/x^2}$, and then let $x\to\infty$ $\endgroup$
    – Empy2
    Nov 12, 2014 at 16:03
  • $\begingroup$ Ohhhh @Michael Got it :D $\endgroup$
    – The Artist
    Nov 12, 2014 at 16:04
  • $\begingroup$ @Ari (+1) Thank you so much :) I understood :) $\endgroup$
    – The Artist
    Nov 12, 2014 at 16:04
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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