How can we prove $$\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}\mathrm{d} x=\frac{2\pi}{3\sqrt 3}?$$

Thought 1
It cannot be solved by using contour integration directly. If we replace $-1/3$ with $-2/3$ or $1/3$ or something else, we can use contour integration directly to solve it.
Thought 2
I have tried substitution $x=t^3$ and $x=1-t$. None of them worked. But I noticed that the form of $1-x+x^2$ does not change while applying $x=1-t$.
Thought 3
Recall the integral representation of $_2F_1$ function, I was able to convert it into a formula with $_2F_1\left(2/3,1;4/3; e^{\pi i/3}\right)$ involved. But I think it will only make the integral more "complex". Moreover, I prefer a elementary approach. (But I also appreciate hypergeometric approach)

  • 1
    $\begingroup$ How did you get that exact answer? Are you sure it's the right one? It might be that $$\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}\mathrm{d} x\neq \frac{2\pi}{3\sqrt 3}$$ $\endgroup$ – Number Feb 24 at 14:15
  • 1
    $\begingroup$ @Zacky Desmos numerically confirms it $\endgroup$ – TheSimpliFire Feb 24 at 14:32
  • 1
    $\begingroup$ If one got any ideas, the integral also equals to $$2^\frac53 \int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$$ $\endgroup$ – Number Feb 24 at 15:49
  • 2
    $\begingroup$ @Zacky In the cosine version of the integral, take $\cos x=\frac12\left(z+\frac1z\right)$ with $z=\exp(ix)\implies dx=\frac{dz}{iz}$ around the circle of radius $\frac\pi2$ centred at $\frac\pi4$. Then you should get a nice quartic for the denominator and two of its roots are simple poles. Use Cauchy's residue formula to complete. $\endgroup$ – TheSimpliFire Feb 24 at 16:47
  • $\begingroup$ @TheSimpliFire I am not familiar with the contour: the circle of radius $\pi/2$ centered at $\pi/4$. I have only encountered the circle which radius is $1$ and is centered at $0$. I tried substitution $z=e^{ix}$ but there seems to be a branch cut in the circle. $\endgroup$ – Kemono Chen Feb 25 at 4:46

The solution heavily exploits symmetry of the integrand.

Let $$I = \int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2} dx $$ Replace $x$ by $1-x$ and sum up gives $$\tag{1} 2I = \int_0^1 \frac{x^{2/3}(1-x)^{-1/3} + (1-x)^{2/3}x^{-1/3}}{1-x+x^2} dx = \int_0^1 \frac{x^{-1/3}(1-x)^{-1/3}}{1-x+x^2} dx$$

Let $\ln_1$ be complex logarithm with branch cut at positive real axis, while $\ln_2$ be the one whose cut is at negative real axis. Denote $$f(z) = \frac{2}{3}\ln_1(x) - \frac{1}{3}\ln_2 (1-x)$$ Then $f(z)$ is discontinuous along the positive axis, but have different jump in $\arg$ across intervals $[0,1]$ and $[1,\infty)$.

Now integrate $g(z) = e^{f(z)}/(1-z+z^2)$ using keyhole contour. Let $\gamma_1$ be path slightly above $[0,1]$, $\gamma_4$ below. $\gamma_2$ be path slightly above $[1,\infty)$, $\gamma_3$ below. It is easily checked that $$\int_{\gamma 1} g(z) dz = I \qquad \qquad \int_{\gamma 4} g(z) dz = I e^{4\pi i/3}$$ $$\int_{\gamma 2} g(z) dz = e^{\pi i/3} \underbrace{\int_1^\infty \frac{x^{2/3}(x-1)^{-1/3}}{1-x+x^2} dx}_J\qquad \int_{\gamma 3} g(z) dz = e^{\pi i} J$$

If we perform $x\mapsto 1/x$ on $J$, we get $\int_0^1 x^{-1/3}(1-x)^{-1/3}/(1-x+x^2)dx$, thus $J = 2I$ by $(1)$.

Therefore $$I(1-e^{4\pi i/3}) + 2I(e^{\pi i / 3} - e^{\pi i}) = 2\pi i\times \text{Sum of residues of } g(z) \text{ at } e^{\pm 2\pi i /3}$$ From which I believe you can work out the value of $I$.

  • 1
    $\begingroup$ (+!) Nice observation on the symmetry of the integrand! $\endgroup$ – Sangchul Lee Feb 25 at 11:12


I have now finally found a way to take my hypergeometric solution all the way to its final elementary form.

Let $$I = \int_0^1 \frac{x^{2/3}}{\sqrt[3]{1 - x} (1 - x +x^2)} \, dx. \tag1$$ Enforcing a substitution of $x \mapsto 1 - x$ leads to $$I = \int_0^1 \frac{(1 - x)^{2/3}}{\sqrt[3]{x} (1 - x + x^2)} \, dx. \tag2$$ Adding (1) to (2) produces $$I = \frac{1}{2} \int_0^1 \frac{dx}{\sqrt[3]{x - x^2} (1 - x + x^2)}.$$

Expanding the second term appearing in the denominator in terms of a geometric series, we have \begin{align} I &= \frac{1}{2} \int_0^1 \frac{dx}{(x - x^2)^{1/3} [1 - (x - x^2)]}\\ &= \frac{1}{2} \int_0^1 \frac{1}{(x - x^2)^{1/3}} \sum_{n = 0}^\infty (x - x^2)^n \, dx\\ &= \frac{1}{2} \sum_{n = 0}^\infty \int_0^1 x^{n - 1/3} (1 - x)^{n - 1/3} \, dx\\ &= \frac{1}{2} \sum_{n = 0}^\infty \operatorname{B} \left (n + \frac{2}{3}, n + \frac{2}{3} \right ), \tag3 \end{align} where $\operatorname{B}(x,y)$ is the Beta function. Making use of the result $$\operatorname{B} (x,x) = \frac{\sqrt{\pi} 2^{1 - 2x} \Gamma (x)}{\Gamma \left (x + \frac{1}{2} \right )},$$ the sum in (3) can be written as \begin{align} I &= \frac{\sqrt{\pi}}{2 \sqrt[3]{2}} \sum_{n = 0}^\infty \frac{\Gamma \left (n + \frac{2}{3} \right )}{\Gamma \left (n + \frac{7}{6} \right ) 4^n}\\ &= \frac{\sqrt{\pi}}{2 \sqrt[3]{2}} \cdot \frac{\Gamma (\frac{2}{3})}{\Gamma (\frac{7}{6})} \sum_{n = 0}^\infty \frac{\left (\frac{2}{3} \right )_n (1)_n}{\left (\frac{7}{6} \right )_n 4^n n!}\\ &= \frac{\sqrt{\pi}}{2 \sqrt[3]{2}} \cdot \frac{\Gamma (\frac{2}{3})}{\Gamma (\frac{7}{6})}\ _2F_1 \left (\frac{2}{3}, 1; \frac{7}{6}; \frac{1}{4} \right ),\tag4 \end{align} where $_2F_1 (a,b;c;x)$ is the Gauss hypergeometric function.

To reduce the hypergeometric function that appears in (4) into elementary form, we proceed as follows.

Firstly, since $_2F_1 (a,b;c;x) =\ _2F_1 (b,a;c;x)$ on applying the second of Pfaff's transformations, namely $$_2F_1 (a,b;c;x) = (1 - x)^{-a}\ _2F_1 \left (a,c-b;c;\frac{x}{x - 1} \right ),$$ to the hypergeometric function, we have $$_2F_1 \left (1, \frac{2}{3}; \frac{7}{6}; \frac{1}{4} \right ) = \frac{4}{3}\ _2F_1 \left (1, \frac{1}{2}; \frac{7}{6}; -\frac{1}{3} \right ).\tag5$$ Next, applying Euler's transformation, namely $$_2F_1 (a,b;c;x) = (1 - x)^{c - a - b}\ _2F_1 (c-a,c-b;c;x),$$ we have $$_2F_1 \left (1, \frac{2}{3}; \frac{7}{6}; \frac{1}{4} \right ) = \frac{4^{2/3}}{3^{2/3}}\ _2F_1 \left (\frac{1}{6}, \frac{2}{3}; \frac{7}{6}; -\frac{1}{3} \right ). \tag6$$

Finally, from DLMF: 15.4.31 we see that $$_2F_1 \left (a, \frac{1}{2} + a; \frac{3}{2}-2a; -\frac{1}{3} \right ) = \left (\frac{8}{9} \right )^{-2a} \frac{\Gamma (\frac{4}{3}) \Gamma (\frac{3}{2} - 2a)}{\Gamma (\frac{3}{2}) \Gamma (\frac{4}{3} - 2a)}.$$ Setting $a = 1/6$ leads to $$_2F_1 \left (\frac{1}{6}, \frac{2}{3}; \frac{7}{6}; -\frac{1}{3} \right ) = \frac{\sqrt[3]{9}}{\sqrt{\pi}} \Gamma \left (\frac{4}{3} \right ) \Gamma \left (\frac{7}{6} \right ).$$ Thus (6) becomes $$_2F_1 \left (1, \frac{2}{3}; \frac{7}{6}; \frac{1}{4} \right ) = \frac{2^{4/3}}{\sqrt{\pi}} \Gamma \left (\frac{4}{3} \right ) \Gamma \left (\frac{7}{6} \right ).$$ On substituting this result into (4), one has \begin{align} I &= \Gamma \left (\frac{2}{3} \right ) \Gamma \left (\frac{4}{3} \right )\\ &= \frac{1}{3} \Gamma \left (\frac{2}{3} \right ) \Gamma \left (\frac{1}{3} \right )\\ &= \frac{1}{3} \Gamma \left (1 - \frac{1}{3} \right ) \Gamma \left (\frac{4}{3} \right )\\ &= \frac{\pi}{3 \sin (\frac{\pi}{3})}\\ &= \frac{2\pi}{3 \sqrt{3}}, \end{align} as required.

  • $\begingroup$ I appreciate your great manipulation of hypergeometric function and your effort researching the special values of $_2F_1$. Anyway, (+1). $\endgroup$ – Kemono Chen Feb 26 at 4:50
  • $\begingroup$ A monumental effort. Well done. $\endgroup$ – Mark Viola Feb 26 at 19:49

There is a completely elementary way to solve this. In the end, I do not see how to find the elementary primitive in a simple and intuitive way (if others do, then please edit the answer accordingly), but Rubi helped me. For this reason I post this but make it cw. I also would like to thank @JanG who pointed me to the question and who actually was the one doing the first changes of variables.

Set $$ I=\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x(1-x)}\,dx. $$ By doing $x\mapsto 1-x$ and adding, others have found that $$ I= \frac{1}{2}\int_0^1\frac{x^{-1/3}(1-x)^{-1/3}}{1-x(1-x)}\,dx. $$ Next, let $x=(1+y)/2$. Then the integral becomes $$ I=2^{2/3}\int_{-1}^1\frac{1}{(1-y^2)^{1/3}(3+y^2)}\,dy= 2^{5/3}\int_{0}^1\frac{1}{(1-y^2)^{1/3}(3+y^2)}\,dy. $$ This is very similar to (and just a $y=\cos t$ away from) the integral @Zacky observes in a comment to the question.

This can be put into Rubi, and surprisinlgy the result is elementary, $$ I=\biggl[\frac{1}{\sqrt{3}}\arctan\Bigl(\frac{\sqrt{3}}{y}\Bigr) +\frac{1}{\sqrt{3}}\arctan\Bigl(\frac{\sqrt{3}\bigl(1-(2-2y^2)^{1/3}\bigr)}{y}\Bigr)-\frac{1}{3}\text{artanh}\,y+\text{artanh}\,\Bigl(\frac{y}{1+(2-2y^2)^{1/3}}\Bigr)\biggr]_0^1 $$ Inserting the boundarys, the upper one gives (using a limit) $2\pi/(3\sqrt{3})$ and the lower one gives $0$. Hence $$ I=\frac{2\pi}{3\sqrt{3}}. $$

  • $\begingroup$ By observing that $\int_0^1\frac{x^{-1/3}(1-x)^{-1/3}}{1-x(1-x)}dx$ has an elementary antiderivative after substitution, the former itself must have too. Huh, Mathematica must have missed something. :) Anyway, it's a great solution, even it only says that the integrand has an elementary antiderivative. $\endgroup$ – Kemono Chen Feb 28 at 8:10

Here we piggy back off the solution posted by @pisco, organize the analysis with detail on the definitions of $\arg(z)$ and $\arg(1-z)$, and finish by evaluating the resiudes enclosed by the closed "keyhole contour."

Let $f(z)$ be the function given by


where choose the branch cut from $0$ to $\infty$ along the positive real axis such that

$$\arg(z)=\begin{cases} 0&, z=x+i0^+\\\\ 2\pi&,z=x+i0^- \end{cases}$$

and we choose the branch cut from $1$ to $\infty$ along the positive real axis with $\arg(1-z)=-\pi+\arg(z-1)$ such that

$$\arg(1-z)=\begin{cases} 0&, 0<x<1\\\\ -\pi&,z=x+i0^+, 1<x\\\\ \pi&, z=x+i0^-, 1<x \end{cases}$$

Then, the integral around the classical "key hole" contour $C$ is

$$\begin{align} \oint_C f(z)\,dz &=(e^{i2(0)/3}e^{-i(0)/3}-e^{i2(2\pi)/3}e^{-i(0)/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\\\\ &+(e^{i2(0)/3}e^{-i(-\pi)/3}-e^{i2(2\pi)/3}e^{-i(\pi)/3})\int_1^\infty \frac{x^{2/3}(x-1)^{-1/3}}{x^2-x+1}\,dx\\\\ &=(1+e^{i\pi/3})\left(\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx+\int_1^\infty \frac{x^{2/3}(x-1)^{-1/3}}{x^2-x+1}\,dx\right)\tag1 \end{align}$$

Enforcing the substitution $x\mapsto 1/x$ in the second integral on the right-hand side of $(1)$ reveals

$$\begin{align} \oint_C f(z)\,dz &=(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}+x^{-1/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\tag2 \end{align}$$

Using the identity $x^{2/3}(1-x)^{-1/3}+x^{-1/3}(1-x)^{2/3}=x^{-1/3}(1-x)^{-1/3}$ and observing that $x^2-x+1=(1-x)^2-(1-x)+1$ we find from $(2)$ that

$$\begin{align} \oint_C f(z)\,dz &=3(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\\\\ &=3(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\tag3 \end{align}$$

From the residue theorem we have

$$\begin{align} \oint_C f(z)\,dz&=2\pi i \left(\text{Res}\left(f(z), z=\frac12+i\frac{\sqrt3}2\right)+\text{Res}\left(f(z), z=\frac12-i\frac{\sqrt3}2\right)\right)\\\\ &=2\pi i \left(\frac{e^{i2\pi/9}e^{i\pi/9}}{i2\sqrt 3}+\frac{e^{i10\pi/9}e^{-i\pi/9}}{-i2\sqrt 3}\right)\\\\ &=\frac{2\pi}{\sqrt3} (1+e^{i\pi/3})\tag4 \end{align}$$

Finally, setting $(3)$ and $(4)$ equal yields the coveted result

$$\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx=\frac{2\pi }{3\sqrt 3}$$

  • $\begingroup$ @pisco I modified your approach a bit and provided a bit more detail to facilitate the presentation to readers who are less familiar with contour integration. I hope that you don't mind. $\endgroup$ – Mark Viola Feb 25 at 18:18
  • $\begingroup$ This answer is great, but I can't accept two answers. (+1) $\endgroup$ – Kemono Chen Feb 25 at 23:44
  • $\begingroup$ @KemonoChen Thank you. And yes, pisco's answer inspired me to post a slightly modified version with more details and carried through to completion. They are effectively the same. $\endgroup$ – Mark Viola Feb 25 at 23:51
  • $\begingroup$ Thank you very much for writing out the details so clearly:). I was in a rush when I typed the solution. $\endgroup$ – pisco Feb 26 at 5:20
  • $\begingroup$ @pisco You're welcome. My pleasure. And thank you for the inspiration. $\endgroup$ – Mark Viola Feb 26 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.