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The question is to evaluate:

$$ \int_{0}^{1}\frac{1}{\sqrt[3]{x^{2}-x^{3}}}\text{d}x $$

using elementary methods, without using complex analysis.

I have a fundamental understanding of integral calculus, including techniques such as substitution, integration by parts, and the evaluation of basic definite and indefinite integrals. The integral above is solved on various sites using advanced tools like complex analysis, which are currently beyond my scope.

My Work:

$$ \text{I} = \int_{0}^{1}\frac{1}{\sqrt[3]{x^{2}-x^{3}}}\text{d}x = \int_{0}^{1}\frac{1}{x \cdot \sqrt[3]{\frac{1}{x} - 1}}\text{d}x $$

$\text{Let}\;\;\dfrac{1}{x} - 1 = t^3 \implies x = \dfrac{1}{1+t^3} \implies \text{d}x = -\dfrac{3t^2}{(1+t^3)^2} \text{d}t $

Substitute these into the integral:

$$ \text{I} = \int_{\infty}^{0} \frac{1+t^3}{t} \cdot \frac{-3t^2}{(1+t^3)^2}\text{d}t = \int_{0}^{\infty} \frac{3t}{1+t^3}\text{d}t $$

This is where my progress ends. I can not further think of any substitution which would do some further progress in evaluating this integral. Any help is appreciated.

dev
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    Hint: since $1+t^3=(1+t)((1+t)^2-3t)$, the integrand is$$\frac{1+t}{1-t+t^2}-\frac{1}{1+t}.$$ – J.G. Dec 08 '24 at 08:49
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    OHHHHH, I'm so dumb, did not expect it to be so easy.. – dev Dec 08 '24 at 08:51
  • Related: https://math.stackexchange.com/questions/397003/integrate-int-01-frac1-sqrt3x2-x3dx?noredirect=1, https://math.stackexchange.com/questions/2367600/using-a-keyhole-contour-for-this-specific-problem?noredirect=1. – Gonçalo Dec 09 '24 at 04:51
  • Lai's answer is much simpler as it exploits the symmetry of the bounds and the denominator under the substitution $x\to\frac1x$, which evades partial fraction decomposition. – Integreek Jan 27 '25 at 15:50

4 Answers4

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$$\int \frac{3t}{1+t^3}dt=\frac{1}{2}\ln(t^2-t+1)-\ln(t+1)+\sqrt{3}\arctan\left({\frac{2t-1}{\sqrt{3}}}\right)$$

Lai
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Delta
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  • That is definitely within my scope. I can’t believe I overlooked something as straightforward as partial fraction decomposition. – dev Dec 08 '24 at 08:59
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Hint

$$\frac{3t}{t^3+1}=\frac{3t}{(t+1)(t^2-t+1)}=\frac{t+1}{t^2-t+1}-\frac{1}{t+1}$$ $$\frac{t+1}{t^2-t+1}=\frac 12 \frac{2t-1}{t^2-t+1}+ \frac 32\,\frac{1}{t^2-t+1}$$

For the last one, complete the square.

Claude Leibovici
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Noting that $$ I=\int_0^{\infty} \frac{3 t}{1+t^3} d t \stackrel{x\mapsto\frac{1}{x}}{=} \int_0^{\infty} \frac{3}{t^3+1} d t $$ Averaging the two versions of the integral $I$ yields $$ \begin{aligned} I & =\frac{3}{2} \int_0^{\infty} \frac{t+1}{t^3+1} d t \\ & =\frac{3}{2} \int_0^{\infty} \frac{1}{t^2-t+1} d t \\ & =6 \int_0^{\infty} \frac{1}{(2 t-1)^2+3} d t\\&=\frac{6}{2 \sqrt{3}}\left[\tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)\right]_0^{\infty} \\ & = \frac {2 \pi}{\sqrt 3} \end{aligned} $$

Wish it helps.

Lai
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Here's the simple non-elementary solution,

$$ I=\int_{0}^{1}\frac{1}{\sqrt[3]{x^{2}-x^{3}}}dx $$

$$ I=\int_{0}^{1} x^{-\frac{2}{3}}(1-x)^{-\frac{1}{3}}\,dx $$

From this,

$${\displaystyle \mathrm {B} (z_{1},z_{2})=\int _{0}^{1}t^{z_{1}-1}(1-t)^{z_{2}-1}\,dt}$$

$$I=B\left(\frac13,\frac23\right)=\frac{2\pi}{\sqrt3}$$

Amrut Ayan
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  • As neat as it is, I'm not sure using the beta function qualifies as elementary method though :-) – zwim Dec 08 '24 at 08:54
  • I'm sorry I don't know about the beta function. – dev Dec 08 '24 at 08:57
  • How do you prove the Gamma function's reflection formula without complex analysis? (If your answer relies on the sine function's Weierstrass product, how do you prove that without complex analysis?) – J.G. Dec 08 '24 at 09:50
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    Hiding Gamma function to undergraduates creates a serious handicap. – Letac Gérard Dec 08 '24 at 14:12
  • @LetacGérard can you please explain your statement? – Integreek May 06 '25 at 11:08
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    In my class of measure theory, I always include one hour for the beta and gamma functions, which are very useful in many circumstances. By the way many exercises of computations by residues can be solved by beta-gamma calculus. To say nothing of statistics classe. – Letac Gérard May 07 '25 at 10:52