14
$\begingroup$

Prove, using elementary methods, that $$\int_0^1 \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}=\frac{4\pi}{3\sqrt 3}$$

I have seen this integral in the following post, however all answers presented exploits complex analysis or heavy series.

But according to mickep's answer even the indefinite integral possess a primitive in terms of elementary functions. I'm not that insane to try and find that by hand, however it gives me great hope that we can find an elementary approach for the definite integral.

Although I kept coming back to it for the past months, I still got no success, or relevant progress and I would appreciate some help.

$\endgroup$
8
  • 2
    $\begingroup$ I would enjoy to see the primitive too ! I tried three different CAS; all of them gave up. $\endgroup$ Jul 5, 2019 at 13:00
  • 1
    $\begingroup$ Do you know Wolfram Cloud ? It is free and you have all Mathemetica. You just need to register. $\endgroup$ Jul 5, 2019 at 13:05
  • 2
    $\begingroup$ just an observation $$\int_{0}^{1}{\frac{2}{(1-x(1-x))}dx}=\frac{4\pi }{3\sqrt{3}}$$ $\endgroup$
    – logo
    Jul 6, 2019 at 21:54
  • 2
    $\begingroup$ And another observation $$\int_0^1 \frac{2}{3 \sqrt[3]{x} (1 - x)^{2/3}} \, dx = \int_0^1 \frac{2}{3\sqrt[3]{x} \sqrt[3]{x(1-x)}} \, dx = \frac{4 \pi}{3 \sqrt{3}}$$ $\endgroup$
    – omegadot
    Jul 8, 2019 at 10:18
  • 3
    $\begingroup$ $f:(0,1)\to\Big(0,\tfrac12\Big),\quad f(x)=\sqrt{x(1-x)}~$ describes a semicircle $($ see geometric mean theorem $),~$ so the fact that $~\displaystyle\int_0^1\frac{dx}{f(x)(1-f^2(x))}~$ depends on $\pi,~$ the constant of the circle, is not exactly a surprise; but the fact that the same holds true even after exchanging square roots for cubic ones is. $\endgroup$
    – Lucian
    Jul 12, 2019 at 21:34

3 Answers 3

21
+50
$\begingroup$

Given the function $f : (0,\,1) \to \mathbb{R}$ $$ f(x) := \frac{1}{\sqrt[3]{x\,(1 - x)}\left(1 - x(1 - x)\right)}\,, $$ we are interested in the calculation of $$ I := \int_0^1 f(x)\,\text{d}x\,. $$ First of all it is good to observe that: $$ f(1 - x) = f(x), \quad \forall \, x \in (0,\,1) $$ then: $$ I = 2 \int_0^{\frac{1}{2}} f(x)\,\text{d}x\,. $$ At this point, since: $$ f\left(\frac{1 - \sqrt{4\,t^3 + 1}}{2}\right) = -\frac{1}{t\left(t^3 + 1\right)}\,, \quad \quad \frac{\text{d}}{\text{d}t}\left(\frac{1 - \sqrt{4\,t^3 + 1}}{2}\right) = -\frac{3\,t^2}{\sqrt{4\,t^3 + 1}} $$ it follows that: $$ I = -6 \int_{-\frac{1}{\sqrt[3]{4}}}^0 \frac{t}{t^3 + 1}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}}\,. $$ Now we can take advantage of the power of Rubi in Wolfram Mathematica:

PacletInstall["https://github.com/RuleBasedIntegration/Rubi/
               releases/download/4.16.1.0/Rubi-4.16.1.0.paclet"];
<< Rubi`
Steps@Int[t/((t^3 + 1) Sqrt[4 t^3 + 1]), t]

enter image description here

enter image description here

from which: $$ I = -6 \int_{-\frac{1}{\sqrt[3]{4}}}^0 \left[ \frac{2\,t - 1}{6\,(t + 1)\,\sqrt{4\,t^3 + 1}} + \frac{t^2}{2\left(t^3 + 1\right)\sqrt{4\,t^3 + 1}} - \\ \frac{2\,t^3 - 3\,t^2 - 1}{6\,t\left(t^2 - t + 1\right)\sqrt{4\,t^3 + 1}} - \frac{1}{6\,t\,\sqrt{4\,t^3 + 1}} \right]\text{d}t\,. $$ Hence a primitive in terms of elementary functions is: $$ I = -6\left[ - \frac{\arctan\left(\frac{\sqrt{3}\,(1 + 2\,t)}{\sqrt{4\,t^3 + 1}}\right)}{3\sqrt{3}} + \frac{\arctan\left(\frac{\sqrt{4\,t^3 + 1}}{\sqrt{3}}\right)}{3\sqrt{3}} - \\ \frac{1}{3}\,\text{arctanh}\left(\frac{1 - 2\,t}{\sqrt{4\,t^3 + 1}}\right) + \frac{1}{9}\,\text{arctanh}\left(\sqrt{4\,t^3 + 1}\right) \right]_{t = -\frac{1}{\sqrt[3]{4}}}^{t = 0} $$ and therefore as desired: $$ I = \frac{4\pi}{3\sqrt{3}}\,. $$


Like any other CAS system, also Rubi follows the rules written by the programmers, so it's always possible to proof by hand how much is executed. Specifically, the theory on which the above rule introduced by Martin Welz is based can be studied in E. GOURSAT Note sur quelques intégrales pseudo-elliptiques (1887). Therefore, based on what is written on page 114, the resolving technique of the integral under consideration can be studied in S. GÜNTHER Sur l’évaluation de certaines intégrales pseudo-elliptiques (1882).

In this case: $$ S \equiv \int \frac{t}{t^3 + 1}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ then imposing: $$ \frac{t}{t^3 + 1} = \frac{\alpha\,t^2}{t^3 + 1} + \frac{\alpha_1\,t^2 + \beta_1\,t + \gamma_1}{t + 1} + \frac{\alpha_2\,t^2 + \beta_2\,t + \gamma_2}{t^2 - t + 1} $$ the identification gives the values: $$ \alpha = \frac{1}{2}\,, \quad \alpha_1 = 0\,, \quad \beta_1 = \frac{1}{3}\,, \quad \gamma_1 = -\frac{1}{6}\,, \quad \alpha_2 = -\frac{1}{3}\,, \quad \beta_2 = \frac{1}{3}\,, \quad \gamma_2 = \frac{1}{6} $$ ie: $$ \frac{t}{t^3 + 1} = \frac{t^2}{2\left(t^3 + 1\right)} + \frac{2\,t - 1}{6\left(t + 1\right)} - \frac{2\,t^2 - 2\,t - 1}{6\left(t^2 - t + 1\right)} $$ from which: $$ S = \int \frac{t^2}{2\left(t^3 + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} + \int \frac{2\,t - 1}{6\left(t + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} + \int \frac{2\,t^2 - 2\,t - 1}{-6\left(t^2 - t + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} \,. $$ Now, for the first integral: $$ S_1 \equiv \int \frac{t^2}{2\left(t^3 + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ according to the method described in the paper: $$ u = \frac{\alpha\,t^3 + \beta\,t^2 + \gamma\,t + \delta}{\sqrt{4\,t^3 + 1}} $$ then imposing: $$ \frac{\text{d}u}{m\,u^2 + n} = \frac{t^2}{2\left(t^3 + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ ie: $$ \frac{\text{d}u}{\text{d}t}\,\frac{2\left(t + 1/t^2\right)\sqrt{4\,t^3 + 1}}{m\,u^2 + n} = 1 $$ the identification gives the values: $$ \alpha = 0\,, \quad \beta = 0\,, \quad \gamma = 0\,, \quad \delta = - \frac{1}{3}\,, \quad m = 27\,, \quad n = 1 $$ ie: $$ S_1 = \int \frac{\text{d}u}{27\,u^2 + 1} = \frac{\arctan\left(3\sqrt{3}\,u\right)}{3\sqrt{3}} + c_1 = -\frac{\arctan\left(\frac{\sqrt{3}}{\sqrt{4\,t^3 + 1}}\right)}{3\sqrt{3}} + c_1\,. $$ Similarly, for the second integral: $$ S_2 \equiv \int \frac{2\,t - 1}{6\left(t + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ putting: $$ u = \frac{\alpha\,t^2 + \beta\,t + \gamma}{\sqrt{4\,t^3 + 1}} $$ then imposing: $$ \frac{\text{d}u}{m\,u^2 + n} = \frac{2\,t - 1}{6\left(t + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ ie: $$ \frac{\text{d}u}{\text{d}t}\,\frac{\frac{6\left(t + 1\right)}{2\,t - 1}\sqrt{4\,t^3 + 1}}{m\,u^2 + n} = 1 $$ the identification gives the values: $$ \alpha = 0\,, \quad \beta = -\frac{2}{3}\,, \quad \gamma = -\frac{1}{3}\,, \quad m = 27\,, \quad n = 1 $$ ie: $$ S_2 = \int \frac{\text{d}u}{27\,u^2 + 1} = \frac{\arctan\left(3\sqrt{3}\,u\right)}{3\sqrt{3}} + c_2 = -\frac{\arctan\left(\frac{\sqrt{3}\left(2\,t + 1\right)}{\sqrt{4\,t^3 + 1}}\right)}{3\sqrt{3}} + c_2\,. $$ For the third integral $$ S_3 \equiv \int \frac{2\,t^2 - 2\,t - 1}{-6\left(t^2 - t + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ this transformation fails and therefore the only hope that remains about the pseudo-ellipticity of the integral consists in further decomposing the rational fraction; in particular, imposing: $$ \frac{2\,t^2 - 2\,t - 1}{-6\left(t^2 - t + 1\right)} = \frac{\alpha}{-6\,t} + \frac{\alpha_1\,t^3 + \beta_1\,t^2 + \gamma_1\,t + \delta_1}{-6\,t\left(t^2 - t + 1\right)} $$ the identification gives the values: $$ \alpha = 1\,, \quad \alpha_1 = 2\,, \quad \beta_1 = -3\,, \quad \gamma_1 = 0\,, \quad \delta_1 = -1 $$ ie: $$ \frac{2\,t^2 - 2\,t - 1}{-6\left(t^2 - t + 1\right)} = \frac{1}{-6\,t} + \frac{2\,t^3 - 3\,t^2 - 1}{-6\,t\left(t^2 - t + 1\right)} $$ from which: $$ S_3 = \int \frac{1}{-6\,t}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} + \int \frac{2\,t^3 - 3\,t^2 - 1}{-6\,t\left(t^2 - t + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} \,. $$ Now, again, for the first integral: $$ S_{3,1} \equiv \int \frac{1}{-6\,t}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ putting: $$ u = \frac{\alpha\,t^3 + \beta\,t^2 + \gamma\,t + \delta}{\sqrt{4\,t^3 + 1}} $$ then imposing: $$ \frac{\text{d}u}{m\,u^2 + n} = \frac{1}{-6\,t}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ ie: $$ \frac{\text{d}u}{\text{d}t}\,\frac{-6\,t\,\sqrt{4\,t^3 + 1}}{m\,u^2 + n} = 1 $$ the identification gives the values: $$ \alpha = 0\,, \quad \beta = 0\,, \quad \gamma = 0\,, \quad \delta = \frac{1}{9}\,, \quad m = -81\,, \quad n = 1 $$ ie: $$ S_{3,1} = \int \frac{\text{d}u}{-81\,u^2 + 1} = \frac{1}{9}\,\text{arctanh}(9\,u) + c_{3,1} = \frac{1}{9}\,\text{arctanh}\left(\frac{1}{\sqrt{4\,t^3 + 1}}\right) + c_{3,1}\,. $$ Finally, for the second integral $$ S_{3,2} \equiv \int \frac{2\,t^3 - 3\,t^2 - 1}{-6\,t\left(t^2 - t + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ putting: $$ u = \frac{\alpha\,t^2 + \beta\,t + \gamma}{\sqrt{4\,t^3 + 1}} $$ then imposing: $$ \frac{\text{d}u}{m\,u^2 + n} = \frac{2\,t^3 - 3\,t^2 - 1}{-6\,t\left(t^2 - t + 1\right)}\,\frac{\text{d}t}{\sqrt{4\,t^3 + 1}} $$ ie: $$ \frac{\text{d}u}{\text{d}t}\,\frac{\frac{-6\,t\left(t^2-t+1\right)}{2\,t^3-3\,t^2-1}\,\sqrt{4\,t^3 + 1}}{m\,u^2 + n} = 1 $$ the identification gives the values: $$ \alpha = 0\,, \quad \beta = \frac{2}{3}\,, \quad \gamma = -\frac{1}{3}\,, \quad m = -9\,, \quad n = 1 $$ ie: $$ S_{3,2} = \int \frac{\text{d}u}{-9\,u^2 + 1} = \frac{1}{3}\,\text{arctanh}(3\,u) + c_{3,2} = \frac{1}{3}\,\text{arctanh}\left(\frac{2\,t - 1}{\sqrt{4\,t^3 + 1}}\right) + c_{3,2}\,. $$ In conclusion, the searched primitive family is $$ S = S_1 + S_2 + S_{3,1} + S_{3,2}\,, $$ which is completely equivalent to that returned by Rubi and therefore evaluating it at the extremes returns what we wanted to prove.

An elementary alternative to avoid the determination of the primitive consists in the parametric method of derivation and integration under the sign of integral (also known as Richard Feynman's trick), but if it isn't possible to identify a winning strategy it's impractical, similar to the method here exposed.

$\endgroup$
4
  • 2
    $\begingroup$ I misunderstood your reason then, thanks for this! And yes, since it exists and can nicely be performed by Rubi it makes me sure that we can also do it by hand. $\endgroup$
    – Zacky
    Jul 5, 2019 at 15:02
  • $\begingroup$ @Zacky the CAS does give a proof of the result though: you can easily verify that identity in red via differentiation. $\endgroup$ Jul 10, 2019 at 16:39
  • $\begingroup$ @Zacky I think "others" obviously agree it's not interesting for a CAS to spit out helpful information. All I was trying to say though is that you do have a complete, elementary proof that you can do by hand at your disposal now. You claim the identity in red, prove it by differentiating, then apply the limits of integration to it. $\endgroup$ Jul 10, 2019 at 16:52
  • 1
    $\begingroup$ Thanks alot for your efforts! I will leave it up for $3$ more days so that people can see your answer. $\endgroup$
    – Zacky
    Jul 12, 2019 at 20:04
4
$\begingroup$

Not elemntary at all for the antiderivative.

Considering $$I=\int \frac{dx}{\sqrt[3]{x(1-x)} (1-x(1-x) )} $$

As Archis Welankar commented, starting with $x=\sin^2(t)$ leads, after simplifcations, to $$I=4 \int\frac{ (1-\cos (4 t))^{2/3} \csc (t) \sec (t)}{7+\cos (4 t)}\,dt$$

Now, $t=\frac{1}{4} \cos ^{-1}(u)$ leads to $$I=-2 \sqrt{2}\int\frac{du}{\sqrt[3]{1-u} \sqrt{u+1} (u+7)}$$ $$I=\frac{12 \sqrt{2}}5 \frac{\sqrt{-u-1}}{(1-u)^{5/6} \sqrt{u+1}}F_1\left(\frac{5}{6};\frac{1}{2},1;\frac{11}{6};-\frac{2}{u-1},-\frac{8}{u-1}\right)$$ where appears the Appell hypergeometric function of two variables.

$\endgroup$
2
  • $\begingroup$ @YuriyS. This is much nicer, for sure ! "Mine" is ugly even after simplifications I forgot to make ! $\endgroup$ Jul 5, 2019 at 14:22
  • $\begingroup$ I made a mistake in latex, it should be $$\int \frac{u^{-1/3} du}{(1-\frac14 u) \sqrt{1-u}}=\frac32 u^{2/3} F_1 \left(\frac23;\frac12, 1 ; \frac53;u, \frac{u}{4} \right)$$ $\endgroup$
    – Yuriy S
    Jul 5, 2019 at 14:29
4
$\begingroup$

Not an answer (yet), just some thoughts.

$$I=\int_0^1 \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}=2 \int_0^{1/2} \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}$$

An obvious substitution:

$$x(1-x)=y$$

$$dx=\frac{dy}{\sqrt{1-4 y}}$$

So we have:

$$I=2 \int_0^{1/4} \frac{y^{-1/3} dy}{(1-y) \sqrt{1-4 y}}$$

Substituting:

$$y=u/4$$

$$I=\frac{4^{1/3}}{2} \int_0^1 \frac{u^{-1/3} du}{(1-\frac14 u) \sqrt{1-u}}$$

This is clearly a hypergeometric function, though it's not considered elementary (which is a pity).

$$I=\frac{4^{1/3}}{2} B \left(\frac{1}{2},\frac{2}{3} \right) {_2 F_1} \left(1,\frac{2}{3}; \frac{7}{6}; \frac{1}{4} \right)$$

I'll continue this in a few hours, the integral seems quite interesting.

Wolfram Alpha can't simplify the above expression to its exact value, which is even more interesting.

A more general form (but not really what the OP wants) would be:

$$I(z)=\int_0^1 \frac{dx}{\sqrt[3]{x(1-x)}(1-z x(1-x))}=\frac{4^{1/3}}{2} B \left(\frac{1}{2},\frac{2}{3} \right) {_2 F_1} \left(1,\frac{2}{3}; \frac{7}{6}; \frac{z}{4} \right)$$


As for the antiderivative, in terms of Appell function we have:

$$I(a)=\int_0^a \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}= \\ =\frac32 (a(1-a))^{2/3} F_1 \left(\frac23; \frac12, 1; \frac53;4 a(1-a), a(1-a) \right) \\ 0 < a < \frac12$$

So far no idea about the elementary form.


In addition.

$$\int_0^1 \frac{u^{-1/3} du}{(1-\frac14 u) \sqrt{1-u}}=\frac43 \int_0^1 \frac{v^{-1/2} dv}{(1+\frac{1}{3} v) (1-v)^{1/3}}$$

Which gives us another hypergeometric form, and another Appell form for the antiderivative.

$$I=I(1)=\frac{2}{3} 4^{1/3} B \left(\frac{1}{2},\frac{2}{3} \right) {_2 F_1} \left(1,\frac{1}{2}; \frac{7}{6}; -\frac{1}{3} \right)$$

Which Wolfram Alpha also can't simplify. I'll see later if Mathematica can do it.

$\endgroup$
6
  • $\begingroup$ According to mickep's answer from that post the antiderivative is: $$\frac{1}{\sqrt{3}}\arctan\Bigl(\frac{\sqrt{3}}{y}\Bigr) +\frac{1}{\sqrt{3}}\arctan\Bigl(\frac{\sqrt{3}\bigl(1-(2-2y^2)^{1/3}\bigr)}{y}\Bigr)-\frac{1}{3}\text{artanh}\,y+\text{artanh}\,\Bigl(\frac{y}{1+(2-2y^2)^{1/3}}\bigg)$$ Where $y=2x-1$. $\endgroup$
    – Zacky
    Jul 5, 2019 at 14:39
  • $\begingroup$ "Not an answer ? A beauty !! $\endgroup$ Jul 5, 2019 at 14:41
  • $\begingroup$ @Zacky, this looks tedious to derive, but knowing the answer should help. If I have the time, I'll come back to this later. Cheers! $\endgroup$
    – Yuriy S
    Jul 5, 2019 at 14:41
  • $\begingroup$ How did you prove $$\int_0^1 \frac{dx}{\sqrt[3]{x(1-x)}(1-z x(1-x))}=\frac{4^{1/3}}{2} \mathrm{B} \left(\frac{1}{2},\frac{2}{3} \right) {_2 F_1} \left(1,\frac{2}{3}; \frac{7}{6}; \frac{z}{4} \right)$$ $\endgroup$
    – clathratus
    Jul 5, 2019 at 16:39
  • 1
    $\begingroup$ I have taken your first expression for $I$ involving the hypergeometric function all the way to its final elementary form in my answer given here. It involves several transformations involving the hypergeometric function and is by no means easy. $\endgroup$
    – omegadot
    Jul 6, 2019 at 2:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .