Playing with integrals of type, $$ I(n)=\int_{-1}^0 \sqrt[2n+1]{x-\sqrt[2n+1]x} \mathrm dx, $$ $$ n \in \mathbb{N} $$ I got two interesting results for the limiting cases $n=1$ and $n \to \infty$: $$ \lim_{n \to \infty} I(n) = 1 $$ The second result is perhaps more interesting, $$ I(1)=\int_{-1}^0 \sqrt[3]{x-\sqrt[3]x} \mathrm dx \approx \frac {\pi}{\sqrt {27}} $$ The approximation is valid to $12$ places of decimal. My question is straight, can we prove these results analytically? Any help would be appreciated.

  • $\begingroup$ You should avoid the ambiguity due to complex roots in using the next integral : $$\int_0^1 \sqrt[2n+1]{x+\sqrt[2n+1]x} \mathrm dx.$$ $\endgroup$ – JJacquelin Apr 1 at 11:38

For $n \in \mathbb{N}$ we have \begin{align} I (n) &= \int \limits_{-1}^0 \left[x - x^{\frac{1}{2n+1}}\right]^{\frac{1}{2n+1}} \mathrm{d} x \stackrel{x = -y}{=} \int \limits_0^1 \left[y^{\frac{1}{2n+1}} - y\right]^{\frac{1}{2n+1}} \mathrm{d} y = \int \limits_0^1 y^{\frac{1}{(2n+1)^2}}\left[1 - y^{\frac{2n}{2n+1}}\right]^{\frac{1}{2n+1}} \mathrm{d} y \\ &\hspace{-10pt}\stackrel{y = t^{\frac{2n+1}{2n}}}{=} \frac{2n+1}{2n} \int \limits_0^1 t^{\frac{n+1}{n (2n+1)}} (1-t)^{\frac{1}{2n+1}} \mathrm{d}t = \frac{2n+1}{2n} \operatorname{B}\left(\frac{n+1}{n(2n+1)}+1,\frac{1}{2n+1} + 1\right) \, . \end{align} Using $\Gamma(x+1) = x \Gamma(x)$ we can rewrite this result to find $$ I (n) = \frac{\operatorname{B} \left(\frac{1}{n} - \frac{1}{2n+1}, \frac{1}{2n+1}\right)}{2 (2n+1)} = \frac{1}{2(2n+1)} \frac{\operatorname{\Gamma}\left(\frac{1}{n} - \frac{1}{2n+1}\right) \operatorname{\Gamma}\left(\frac{1}{2n+1}\right)}{\operatorname{\Gamma}\left(\frac{1}{n}\right)}$$ for $n \in \mathbb{N}$. In particular, $$ I(1) = \frac{\operatorname{\Gamma}\left(\frac{2}{3}\right) \operatorname{\Gamma}\left(\frac{1}{3}\right)}{6} = \frac{\pi}{6 \sin \left(\frac{\pi}{3}\right)} = \frac{\pi}{3 \sqrt{3}} \,.$$ Moreover, we obtain $$ \lim_{n \to \infty} I (n) \stackrel{\Gamma(x) \, \stackrel{x \to 0}{\sim} \, \frac{1}{x}}{=} \lim_{n \to \infty} \frac{1}{2(2n+1)} \frac{\frac{n(2n+1)}{n+1} (2n+1)}{n} = \lim_{n \to \infty} \frac{2n+1}{2(n+1)} = 1 \, .$$

  • $\begingroup$ I think that your calculus is not correct. At first line you supposed $x^{\frac{2n}{2n+1}}= (-x)^{\frac{2n}{2n+1}}=y^{\frac{2n}{2n+1}}$ which is false. It is true that $((-x)^{2})^{\frac{n}{2n+1}}=y^{\frac{2n}{2n+1}}$ but $ x^{\frac{2n}{2n+1}}\neq ((-x)^{2})^{\frac{n}{2n+1}}$. The first is complex. The second is real. $\endgroup$ – JJacquelin Apr 1 at 11:02
  • $\begingroup$ Interesting because I get another solution but I can not reply the second result $\endgroup$ – stocha Apr 1 at 11:05
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    $\begingroup$ @JJacquelin I have added an intermediate step to illustrate what I have done here. The only assumption is that $x^{1/(2n+1)} = - (-x)^{1/(2n+1)}$ holds for $x < 0$, which (as confirmed by the results) seems to agree with the OP's definition of odd roots of negative numbers. $\endgroup$ – ComplexYetTrivial Apr 1 at 11:10
  • $\begingroup$ @ComplexYetTrivial. I agree that the key point is the definition of odd roots of negative numbers. $\sqrt[3]{-1}$ has three roots, $−1$ and two complex. I would agree with the OP results if the integral was $$\int_0^1\sqrt[2n+1]{x+\sqrt[2n+1]{x} }dx $$ $\endgroup$ – JJacquelin Apr 1 at 11:54
  • $\begingroup$ @JJacquelin, I am truly sorry if it is sarcastic, but I think at this point Wolfram Alpha has got a bug, for the integrand is indeed defined for all real $x$, whether positive or negative . In fact you can manually plot it on both sides of the y-axis and find that there is indeed a positive area bound by the curve and the y-axis between -1 and. 0. $\endgroup$ – Awe Kumar Jha Apr 1 at 13:14

Too long for a comment.

After ComplexYetTrivial's answer, we have $$I_n=\frac{\Gamma \left(\frac{2 (n+1)}{2 n+1}\right) \Gamma \left(\frac{n+1}{n(2 n+1)}\right)}{2 \Gamma \left(\frac{1}{n}\right)}$$ Expanding as series $$I_n=1-\frac{1}{2 n}+\frac{12-\pi ^2}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ which is not "too bad" even for $n=1$; this would give $1-\frac{\pi ^2}{24}\approx 0.588766$ while $\frac{\pi}{3 \sqrt{3}}\approx 0.604600$.

For a few values of $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 1 & 0.588766 & 0.604600 \\ 2 & 0.772192 & 0.774848 \\ 3 & 0.843196 & 0.843965 \\ 4 & 0.880548 & 0.880851 \\ 5 & 0.903551 & 0.903695 \\ 6 & 0.919132 & 0.919210 \\ 7 & 0.930383 & 0.930429 \\ 8 & 0.938887 & 0.938916 \\ 9 & 0.945540 & 0.945560 \end{array} \right)$$

  • $\begingroup$ well I didn't think that the 2nd degree Taylor approximation would be so accurate, +1. $\endgroup$ – Awe Kumar Jha Apr 3 at 11:31
  • $\begingroup$ @AweKumarJha. Me neither, be sure ! In fact, we can easily go further. Cheers :-) $\endgroup$ – Claude Leibovici Apr 3 at 16:19
  • $\begingroup$ @AweKumarJha. Still better if we use $$I_n=1-\frac{1}{2 n}+\frac{12-\pi ^2}{24 n^2}+\frac{12 (\zeta (3)-2)+\pi ^2}{48 n^3}+O\left(\frac{1}{n^4}\right)$$ The numbers would become $$\{0.594897,0.772958,0.843423,0.880644,0.903600,0.919161,0.930401,0.938899,0.945549\}$$ $\endgroup$ – Claude Leibovici Apr 4 at 4:42

I found the solution of your interesting integral by try:

$$\int_{-1}^0 \left(x-x^{\frac{1}{2 n+1}}\right)^{\frac{1}{2 n+1}} \,dx = \frac{(-1)^{\frac{2 n}{1+2 n}} \left(-1+(-1)^{1+\frac{1}{1+2 n}}\right)^{1+\frac{1}{1+2 n}} (1+2 n)^2 \text{Hypergeometric2F1}\left[1,2+\frac{1}{n},2+\frac{1}{n}-\frac{1}{1+2n},(-1)^{\frac{2 n}{1+2 n}}\right]}{2 (1+2 n (1+n))}$$

I checked the result for several values, but since I can't get your second result, the numerical solution is complex, I guess you did a mistake in your post! Please check your second result!


I am afraid that your calculus is not correct. Unfortunately the calculus for $I(1)$ is not detailed. One cannot say where exactly is the mistake.

I guess that the trouble comes from a transformation such as $x^{\frac{2n}{2n+1}}= ((-x)^{2})^{\frac{n}{2n+1}}$ which is false. $$ x^{\frac{2n}{2n+1}}\neq ((-x)^{2})^{\frac{n}{2n+1}}$$ for $-1<x<0$ the first term is complex. The second is real.

The integral is not real $\simeq \frac{\pi}{\sqrt{27}}$. $$I(1)=\int_{-1}^0 \sqrt[3]{x-\sqrt[3]x} \mathrm dx \simeq 0.68575-0.242772\,i$$

  • $\begingroup$ Yes your right, see my post below, I recognize the same $\endgroup$ – stocha Apr 1 at 11:30

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