5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $$ \lim_{n \to \infty} I (n) \stackrel{\Gamma(x) \, \stackrel{x \to 0}{\sim} \, \frac{1}{x}}{=} \lim_{n \to \infty} \frac{1}{2(2n+1)} \frac{\frac{n(2n+1)}{n+1} (2n+1)}{n} = \lim_{n \to \infty} \frac{2n+1}{2(n+1)} = 1 \, .$$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $\simeq \frac{\pi}{\sqrt{27}}$. $$I(1)=\int_{-1}^0 \sqrt[3]{x-\sqrt[3]x} \mathrm dx \simeq 0.68575-0.242772\,i$$

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

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.