Vers la Prépa (4) - Exercice 1

On a :
$$\mathcal{I} = \int_{0}^{1} \sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}}} \, dx$$
Or, pour $$X \geqslant 0$$, on sait que $$\sqrt{X} = X^\frac{1}{2}$$. Ce qui va nous donner :
$$\mathcal{I} = \int_{0}^{1} \left(x\left(x\left(x\left(x\left(x\left( x \right)^\frac{1}{2}\right)^\frac{1}{2}\right)^\frac{1}{2}\right)^\frac{1}{2}\right)^\frac{1}{2} \right)^\frac{1}{2} \, dx$$
Mais, on a $$XX^\frac{1}{2} =X^1X^\frac{1}{2} =X^{1+\frac{1}{2}}= X^\frac{3}{2}$$ Ce qui va nous permettre d'écrire que :
$$\mathcal{I} = \int_{0}^{1} <br />\left(x\left(x\left(x\left(x\left(x^\frac{3}{2}\right)^\frac{1}{2}\right)^\frac{1}{2}\right)^\frac{1}{2}\right)^\frac{1}{2} \right)^\frac{1}{2} \, dx$$
Soit encore :
$$\mathcal{I} = \int_{0}^{1} \left(x\left(x\left(x\left(xx^\frac{3}{4} \right)^\frac{1}{2}\right)^\frac{1}{2}\right)^\frac{1}{2} \right)^\frac{1}{2} \, dx \,\,\,\,\, \Longleftrightarrow \,\,\,\,\, \mathcal{I} = \int_{0}^{1} \left(x\left(x\left(x\left(x^\frac{7}{4} \right)^\frac{1}{2}\right)^\frac{1}{2}\right)^\frac{1}{2} \right)^\frac{1}{2} \, dx$$
Donc, on obtient :
$$\mathcal{I} = \int_{0}^{1} \left(x\left(x\left(xx^\frac{7}{8}\right)^\frac{1}{2}\right)^\frac{1}{2} \right)^\frac{1}{2} \, dx \,\,\,\,\, \Longleftrightarrow \,\,\,\,\, \mathcal{I} = \int_{0}^{1} \left(x\left(x\left(x^\frac{15}{8}\right)^\frac{1}{2}\right)^\frac{1}{2} \right)^\frac{1}{2} \, dx$$
C'est-à-dire :
$$\mathcal{I} = \int_{0}^{1} \left(x\left(x x^\frac{15}{16}\right)^\frac{1}{2} \right)^\frac{1}{2} \, dx \,\,\,\,\, \Longleftrightarrow \,\,\,\,\, \mathcal{I} = \int_{0}^{1} \left(x\left(x^\frac{31}{16} \right)^\frac{1}{2} \right)^\frac{1}{2} \, dx$$
Dès lors, on trouve que :
$$\mathcal{I} = \int_{0}^{1} \left(x x^\frac{31}{32}\right)^\frac{1}{2} \, dx\,\,\,\,\, \Longleftrightarrow \,\,\,\,\,\mathcal{I} = \int_{0}^{1} \left(x^\frac{63}{32}\right)^\frac{1}{2} \, dx \,\,\,\,\, \Longleftrightarrow \,\,\,\,\, \mathcal{I} = \int_{0}^{1} x^\frac{63}{64} \, dx$$
En intégrant, on obtient alors :
$$\mathcal{I} = \left[ \dfrac{x^{\frac{63}{64}+1}}{\dfrac{63}{64}+1}\right]_{0}^{1} \,\,\,\,\, \Longleftrightarrow \,\,\,\,\, \mathcal{I} = \left[ \dfrac{x^{\frac{127}{64}}}{\dfrac{127}{64}}\right]_{0}^{1}\,\,\,\,\, \Longleftrightarrow \,\,\,\,\, \mathcal{I} = \dfrac{64}{127} \left[ x^{\frac{127}{64}} \right]_{0}^{1}$$
Finalement :
$$\boxed{\mathcal{I} = \dfrac{64}{127} \, u.a.}$$