Dgdrxrt

Playing with integrals of type,
$$
I(n)=int_-1^0 sqrt[2n+1]x-sqrt[2n+1]x mathrm dx,
$$
$$
n in mathbbN
$$
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,
$$
I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx approx frac pisqrt 27
$$
The approximation is valid to $12$ places of decimal.
My question is straight, can we prove these results analytically? Any help would be appreciated.

For $n in mathbbN$ we have
beginalign
I (n) &= int limits_-1^0 left[x - x^frac12n+1right]^frac12n+1 mathrmd x stackrelx = -y= int limits_0^1 left[y^frac12n+1 - yright]^frac12n+1 mathrmd y = int limits_0^1 y^frac1(2n+1)^2left[1 - y^frac2n2n+1right]^frac12n+1 mathrmd y \
&hspace-10ptstackrely = t^frac2n+12n= frac2n+12n int limits_0^1 t^fracn+1n (2n+1) (1-t)^frac12n+1 mathrmdt = frac2n+12n operatornameBleft(fracn+1n(2n+1)+1,frac12n+1 + 1right) , .
endalign
Using $Gamma(x+1) = x Gamma(x)$ we can rewrite this result to find
$$ I (n) = fracoperatornameB left(frac1n - frac12n+1, frac12n+1right)2 (2n+1) = frac12(2n+1) fracoperatornameGammaleft(frac1n - frac12n+1right) operatornameGammaleft(frac12n+1right)operatornameGammaleft(frac1nright)$$
for $n in mathbbN$. In particular,
$$ I(1) = fracoperatornameGammaleft(frac23right) operatornameGammaleft(frac13right)6 = fracpi6 sin left(fracpi3right) = fracpi3 sqrt3 ,.$$
Moreover, we obtain
$$ lim_n to infty I (n) stackrelGamma(x) , stackrelx to 0sim , frac1x= lim_n to infty frac12(2n+1) fracfracn(2n+1)n+1 (2n+1)n = lim_n to infty frac2n+12(n+1) = 1 , .$$

Too long for a comment.

After ComplexYetTrivial's answer, we have
$$I_n=fracGamma left(frac2 (n+1)2 n+1right) Gamma left(fracn+1n(2
n+1)right)2 Gamma left(frac1nright)$$ Expanding as series
$$I_n=1-frac12 n+frac12-pi ^224 n^2+Oleft(frac1n^3right)$$ which is not "too bad" even for $n=1$; this would give $1-fracpi ^224approx 0.588766$ while $fracpi3 sqrt3approx 0.604600$.

For a few values of $n$
$$left(
beginarrayccc
n & textapproximation & textexact \
1 & 0.588766 & 0.604600 \
2 & 0.772192 & 0.774848 \
3 & 0.843196 & 0.843965 \
4 & 0.880548 & 0.880851 \
5 & 0.903551 & 0.903695 \
6 & 0.919132 & 0.919210 \
7 & 0.930383 & 0.930429 \
8 & 0.938887 & 0.938916 \
9 & 0.945540 & 0.945560
endarray
right)$$

I found the solution of your interesting integral by try:

$$int_-1^0 left(x-x^frac12 n+1right)^frac12 n+1 ,dx = frac(-1)^frac2 n1+2 n left(-1+(-1)^1+frac11+2 nright)^1+frac11+2 n (1+2 n)^2 textHypergeometric2F1left[1,2+frac1n,2+frac1n-frac11+2n,(-1)^frac2 n1+2 nright]2 (1+2 n (1+n))$$

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 $x^frac2n2n+1= ((-x)^2)^fracn2n+1$ which is false. $$ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$$ for $-1<x<0$ the first term is complex. The second is real.

The integral is not real $simeq fracpisqrt27$.
$$I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx simeq 0.68575-0.242772,i$$