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

So far I have found the indefinite integral, which is:

$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$

Now what do I need to do in order to calculate the improper integral?

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  • $\begingroup$ You better group the first and last term before taking limits. $\endgroup$
    – Lai
    Dec 7, 2021 at 0:37

6 Answers 6

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As this one has been solved using a keyhole contour I thought I would show that it can also be done using a slice contour. Let $$f(z) = \frac{1}{1+z^3}.$$

The slice consists of three parts parameterized by $R$, which is real and goes to infinity. The first part $\Gamma_1$ is a line going from zero to $R$ along the real axis. The next one is a counterclockwise arc $\Gamma_2$ going from $R$ to $Re^{2i\pi/3}.$ The last one namely $\Gamma_3$ is a line back to the origin from $Re^{2i\pi/3}.$ along the ray at angle $2i\pi/3.$

Now we have the following bound on the integral along the arc: $$\left|\int_{\Gamma_2} f(z) dz\right| \le 2\pi R/3 \frac{1}{R^3-1} \in O(1/R^2) \rightarrow 0 \quad \text{as} \quad R \rightarrow\infty.$$

Moreover along $\Gamma_3$ setting $z = te^{2i\pi/3}$ we have $$\int_{\Gamma_3} f(z) dz = \int_R^0 \frac{1}{1+t^3 e^{2i\pi}} e^{2i\pi/3} dt = - e^{2i\pi/3} \int_0^R \frac{1}{1+t^3} dt.$$

Now with $Q$ being the integral we are looking for, we thus have in the limit $$\int_{\Gamma_1} f(z) dz = Q \quad \text{and}\quad \int_{\Gamma_3} f(z) dz = - e^{2i\pi/3} Q.$$

Applying the Cauchy residue theorem to the slice contour, we obtain $$ Q (1 - e^{2i\pi/3} ) = 2\pi i \operatorname{Res}(f(z); z = e^{i\pi/3}) $$ where $$\operatorname{Res}(f(z); z = e^{i\pi/3}) = \lim_{z\to e^{i\pi/3}} \frac{z-e^{i\pi/3}}{1+z^3} = \lim_{z\to e^{i\pi/3}} \frac{1}{3z^2} =\frac{1}{3} e^{- 2i\pi/3}.$$ It follows that $$ Q= \frac{1}{3} 2\pi i \frac{e^{- 2i\pi/3}}{1 - e^{2i\pi/3}} = \frac{1}{3}2\pi i \frac{e^{- 3i\pi/3}}{e^{-i\pi/3} - e^{i\pi/3}} = \frac{1}{3}\pi \frac{1}{\sin(\pi/3)} = \frac{1}{3}\pi \frac{2}{\sqrt 3} = \frac{2\pi}{3\sqrt{3}}.$$

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  • $\begingroup$ This was great, I appreciate you using the slice contour instead of the keyhole contour which is the only way I've seen this being done. I had a question about your contour integral/residue theorem. Your entire contour also includes the portion $\Gamma_3$, but doesn't $f$ have a pole here (namely: $e^{\frac{2 \pi i}{3}}$)? How can we still use the residue theorem since we have a pole on the contour? Is there something I'm missing? $\endgroup$
    – Mohabat
    Dec 30, 2020 at 19:26
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    $\begingroup$ I get $1/(1+e^{(2\pi i/3)\times 3}) = 1/2.$ $\endgroup$ Dec 30, 2020 at 23:12
  • $\begingroup$ You're completely right, my silly mistake. Thank you! $\endgroup$
    – Mohabat
    Dec 31, 2020 at 14:22
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Next, simplify $$ F(x)=-\frac{1}{6}\ln|x^2-x+1|+\frac{1}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}+\frac{1}{3}\ln|x+1| $$ $$ =\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|-\frac{1}{3}\ln\sqrt{|x^2-x+1|} $$ $$ =\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln\left(\frac{|x+1|}{\sqrt{|x^2-x+1|}}\right). $$ Then $$\int_0^\infty \frac{dx}{1+x^3}=\lim_{X\rightarrow\infty}F(X)-F(0).$$ Compute the limit, and you are done.

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  • $\begingroup$ Got it. Thanks. Missed the last part in class :-) $\endgroup$ Apr 11, 2013 at 13:08
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Another method!

$$t=\frac{1}{1+x^3}:$$

$$\begin{aligned}\int_0^{\infty} \frac{dx}{1+x^3}&=\frac{1}{3}\int_0^1 \frac{1}{t}\left(\frac{1}{t}-1\right)^{-2/3} dt\\[7pt]&=\frac{1}{3}\int_0^1 t^{-1/3}\left(1-t\right)^{-2/3}\,dt\\[7pt]&= \frac{1}{3}\text{B}\left(\frac{2}{3},\frac{1}{3}\right)=\frac{1}{3}\Gamma\left(1-\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right)\\[7pt]&=\frac{\pi}{3}\csc\frac{\pi}{3}\\[7pt]&=\frac{2\pi}{3\sqrt{3}}\end{aligned}$$

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To do the improper integral, it is actually easier to use the Residue theorem. In this case, however, you would revert to one of the trickier formulations. That is, consider the integral in the complex plane

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

where $C$ is a keyhole contour about the positive real axis. You may then show that

$$-i 2 \pi \int_0^{\infty} \frac{dx}{1+x^3} = i 2 \pi \sum \text{Res}_{z=z_k} \frac{\log{z}}{1+z^3}$$

The RHS includes the sum of the residues at the poles of the integrand. In this case, the poles are at $z=e^{i \pi/3}$, $z=e^{i \pi}$, and $z=e^{i 5 \pi/3}$. The sum of the residues then becomes

$$ \frac{i \pi/3}{3 e^{i 2 \pi/3}}+\frac{i \pi}{3}+\frac{i 5 \pi/3}{3 e^{-i 2 \pi/3}}= -\frac{2 \pi}{3 \sqrt{3}} $$

The integral is then the negative of this, i.e.,

$$\int_0^{\infty} \frac{dx}{1+x^3} = \frac{2 \pi}{3 \sqrt{3}} $$

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  • $\begingroup$ Well, Seems like a really nice solution. Might study that Residue theorem. Thanks. $\endgroup$ Apr 11, 2013 at 13:10
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    $\begingroup$ Knowledge of this theorem of the techniques of using allows one to solve a whole lot of problems with little fuss. Not sure where to steer you if you are starting from scratch, i.e., Calc II. I used Churchill & Brown undergrand and Ahlfors in the grad course; I think Ahlfors is better. $\endgroup$
    – Ron Gordon
    Apr 11, 2013 at 13:13
  • $\begingroup$ I'm trying to confirm your equation: $$\frac {\pi i}{9\exp(2\pi i/3)}+\frac{\pi i}{3}+\frac{5\pi i}{9\exp(-2\pi i/3)}=-\frac{2\pi}{3\sqrt 3}$$ but either by hand or by calculator LHS has always imaginary part, too. $\endgroup$
    – J. Doe
    Jul 7, 2019 at 13:33
  • $\begingroup$ The calculation in a calculator is: symbolab.com/solver/complex-numbers-calculator/… $\endgroup$
    – J. Doe
    Jul 7, 2019 at 13:41
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Computing a indefinite integral can be harder (or more tedious because of a complicated antiderivative) than computing a definite integral.

In this case, we can do something simpler.

Let $$I = \int_{0}^{\infty} \frac{\text{d}x}{1 + x^3}$$

Make the substitution $x = \dfrac{1}{t}$ (this steps needs to be justified), and we get

$$I = \int_{0}^{\infty} \frac{t}{1 + t^3}\text{d}t$$

Adding up, we get

$$2I = \int_{0}^{\infty} \frac{1+x}{1 + x^3} \text{d}x = \int_{0}^{\infty} \frac{1}{x^2 - x +1} \text{d}x$$

$$ = \int_{0}^{\infty} \frac{1}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \text{d}x$$

which can be computed easily (in terms of the antiderivative!).

(From an earlier answer here, where this was a sub-step in a slightly harder integral: Simpler way to compute a definite integral without resorting to partial fractions?)

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  • $\begingroup$ Nice, but you forgot a $t$ in the second integral. $\endgroup$
    – Mårten W
    Apr 11, 2013 at 13:44
  • $\begingroup$ @MårtenW: You are right, thanks! Fixed. $\endgroup$
    – Aryabhata
    Apr 11, 2013 at 13:49
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$\displaystyle \because \int\dfrac { 1 } { 1 + x ^ { 3 } } d x{\displaystyle =\int\dfrac { 1 - x ^ { 2 } + x ^ { 2 } } { 1 + x ^ { 3 } } d x } $ $ {\displaystyle =\int \dfrac { 1 - x } { 1 - x + x ^ { 2 } } d x + \dfrac { 1 } { 3 } \int \dfrac { d ( 1 + x ^ { 3 } ) } { 1 + x ^ { 3 } } }$

$\displaystyle =-\dfrac { 1 } { 2 } \int \dfrac { d ( 1 - x + x ^ { 2 } ) } { 1 - x + x ^ { 2 } } + \dfrac { 1 } { 2 } \int \dfrac { d x } { ( x - \dfrac { 1 } { 2 } ) ^ { 2 } + ( \dfrac { \sqrt { 3 } } { 2 } ) ^ { 2 } } + \dfrac { 1 } { 3 } \ln | 1 + x ^ { 3 } | $

$ { \displaystyle =\dfrac { 1 } { 6 } \ln \left( \dfrac { ( 1 + x ) ^ { 2 } } { 1 - x + x ^ { 2 } }\right ) + \dfrac { 1 } { \sqrt { 3 } } \tan ^ { - 1 } \left( \dfrac { 2 x - 1 } { \sqrt { 3 } }\right) + C }$

$\therefore \begin{aligned} \int_{0}^{\infty} \frac{1}{1+x^{3}} d x &=\frac{1}{6} \left[\ln\frac{(1+x)^{2}}{1-x+x^{2}}\right]_{0}^{\infty}+\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)\right]_0^\infty=\frac{2 \pi}{3 \sqrt{3}} \end{aligned}\quad \blacksquare$

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