On the integrals $\int_{-1}^0 \sqrt[2n+1]{x-\sqrt[2n+1]x} \mathrm dx$ 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.
 A: 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
$$ \lim_{n \to \infty} I (n) \stackrel{\Gamma(x) \,  \stackrel{x \to 0}{\sim} \, \frac{1}{x}}{=} \lim_{n \to \infty} \frac{1}{2(2n+1)} \frac{\frac{n(2n+1)}{n+1} (2n+1)}{n} = \lim_{n \to \infty} \frac{2n+1}{2(n+1)} = 1 \, .$$ 
A: 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 $n$
$$\left(
\begin{array}{ccc}
n & \text{approximation} & \text{exact} \\
 1 & 0.588766 & 0.604600 \\
 2 & 0.772192 & 0.774848 \\
 3 & 0.843196 & 0.843965 \\
 4 & 0.880548 & 0.880851 \\
 5 & 0.903551 & 0.903695 \\
 6 & 0.919132 & 0.919210 \\
 7 & 0.930383 & 0.930429 \\
 8 & 0.938887 & 0.938916 \\
 9 & 0.945540 & 0.945560
\end{array}
\right)$$
A: 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!
A: 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 $\simeq \frac{\pi}{\sqrt{27}}$.
$$I(1)=\int_{-1}^0 \sqrt[3]{x-\sqrt[3]x} \mathrm dx \simeq 0.68575-0.242772\,i$$
