Solve this Integral for $a$ where $a$ is a real number

$$\large{\int_0^1 ((1-x^a)^\frac{1}{a}-x)^2 dx}$$

can anybody solve this step by step please?

Solutions Collecting From Web of "Solve this Integral for $a$ where $a$ is a real number"

It is useful to recall that, for any $a,b>0$
$$ \int_{0}^{1} u^{a-1}(1-u)^{b-1}\,du = B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.\tag{1}$$

Assuming $a\in\mathbb{R}^+$, that gives $ I = I_1 + I_2 + I_3 $, where:

$$ I_1 = \int_{0}^{1}(1-x^a)^{\frac{2}{a}}\,dx = \frac{1}{a}\int_{0}^{1}u^{\frac{1}{a}-1}(1-u)^{\frac{2}{a}}\,du =\frac{\Gamma\left(\frac{1}{a}\right)\Gamma\left(\frac{a+2}{a}\right)}{a\,\Gamma\left(\frac{a+3}{a}\right)},$$
$$ I_2 = -2\int_{0}^{1} x(1-x^a)^{\frac{1}{a}}\,dx = -\frac{2}{a}\int_{0}^{1}u^{\frac{2}{a}-1}(1-u)^{\frac{1}{a}}\,du = -2\,\frac{\Gamma\left(\frac{2}{a}\right)\Gamma\left(\frac{a+1}{a}\right)}{a\,\Gamma\left(\frac{a+3}{a}\right)}$$
$$ I_3 = \int_{0}^{1}x^2\,dx = \frac{1}{3} \tag{2}$$
and by using $\Gamma(z+1)=z\,\Gamma(z)$ we get:

$$ I = \frac{2}{3a}\left(\frac{\Gamma\left(\frac{1}{a}\right)\Gamma\left(\frac{2}{a}\right)}{\Gamma\left(\frac{3}{a}\right)}-\frac{\Gamma\left(\frac{2}{a}\right)\Gamma\left(\frac{1}{a}\right)}{\Gamma\left(\frac{3}{a}\right)}\right)+\frac{1}{3}=\bbox[5px,border:2px solid #C0A000]{\color{red}{\frac{1}{3}}}\tag{3}$$

Probably we missed some straightforward symmetry, but in any case the answer does not depend on $a$.


We can check that also without using the Beta machinery. Let $f_a(x)=(1-x^a)^{\frac{1}{a}}$.

Through the substitution $x=(1-y^a)^{1/a}$ we have:

$$ I = \int_{0}^{1} (f_a(x)-x)^2\,dx = -\int_{0}^{1}f_a'(y) (f_a(y)-y)^2\,dy \tag{4}$$
hence:
$$ -2I = \int_{0}^{1}(f_a'(y)-1)(f_a(y)-y)^2\,dy = \frac{1}{3}\left.\left(f_a(y)-y)\right)^3\right|_{0}^{1}=-\frac{2}{3}\tag{5}$$
from which $I=\frac{1}{3}$ readily follows.

This proves that it is enough to exploit the fact that $f_a(x)$ is an involutive map, and also proves that:

$$ \forall a\in\mathbb{R}^+,n\in\mathbb{N},\qquad \int_{0}^{1}\left(\left(1-x^a\right)^{\frac{1}{a}}-x\right)^{2n}\,dx = \frac{1}{2n+1}. \tag{6}$$

We have

$$\int_0^1\left((1-x^a)^{1/a}-x\right)^2dx=\int_0^1(1-x^a)^{2/a}dx-2\int_0^1(1-x^a)^{2/a}dx+\int_0^1x^2\,dx \tag 1$$

For the first and second integrals on the right-hand side of $(1)$, let $x=t^{1/a}$ so that $dx=\frac1a t^{1/a-1}$. Then, we have

$$\begin{align}
\int_0^1(1-x^a)^{2/a}dx&=\frac1a\int_0^1t^{1/a-1}(1-t)^{2/a}dt\\\\
&=\frac1a B\left(\frac1a,1+\frac2a\right) \tag 2
\end{align}$$

$$\begin{align}
\int_0^1x(1-x^a)^{1/a}dx&=\frac1a\int_0^1t^{2/a-1}(1-t)^{1/a}dt\\\\
&=\frac1a B\left(\frac2a,1+\frac1a\right) \tag3
\end{align}$$

where in $(2)$ and $(3)$, $B(x,y)$ is the Beta Function defined by

$$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt$$

Finally, using $(2)$ and $(3)$ in $(1)$ gives

$$\int_0^1\left((1-x^a)^{1/a}-x\right)^2dx=\frac13+\frac1a\left( B\left(\frac1a,1+\frac2a\right)-2B\left(\frac2a,1+\frac1a\right)\right)$$

Using $B(x,1+y)=B(x,y)\frac{y}{x+y}$, it is easy to see that the terms involving the Beta Function cancel and we are left with

$$\bbox[5px,border:2px solid #C0A000]{\int_0^1\left((1-x^a)^{1/a}-x\right)^2dx=\frac13}$$

which agrees with the answer provided by @JackD’Aurizio!