Calculus 5 : Interesting radicals

Let us first consider if the following limit exists:

$\begin{aligned} y & = \sqrt {x \sqrt {x \sqrt {x\sqrt {x...}}}} \\ & = x^\frac 12 \times x^\frac 14 \times x^\frac 18 ... \\ & = \lim_{n \to \infty} \prod_{k=1}^n x^\frac 1{2^k} \\ & = \lim_{n \to \infty} \exp \left(\sum_{k=1}^n \frac 1{2^k} \ln x \right) \\ & = \lim_{n \to \infty} \exp \left(\ln x \sum_{k=1}^n \frac 1{2^k} \right) \\ & = \exp \left(\ln x \times \frac {\frac 12}{1-\frac 12} \right) \\ & = e^{\ln x} = x \end{aligned}$

Therefore,

$\begin{aligned} I & = \int_0^{\sqrt{10}} \sqrt {x \sqrt {x \sqrt {x\sqrt {x...}}}} \ dx \\ & = \int_0^{\sqrt{10}} x \ dx \\ & = \frac {x^2}2 \bigg|_0^{\sqrt{10}} \\ & = 5 \end{aligned}$

Relevant wiki: $u$ -Substitution

This actually is a very easy problem, made tricky.

$\displaystyle I = \int_0^{\sqrt[]{10}} ~ \sqrt[]{x\sqrt[]{x\sqrt[]{x\sqrt[]{x\ldots}}}} ~ dx \\ \displaystyle \text{Let } i = \sqrt[]{x\sqrt[]{x\sqrt[]{x\sqrt[]{x\ldots}}}} \\ \displaystyle i^2 = x\sqrt[]{x\sqrt[]{x\sqrt[]{x\ldots}}} = xi \\ \displaystyle \implies x = i \\ \displaystyle \dfrac{di}{dx} = 1 \\ \displaystyle dx = di \\ \displaystyle I = \int_0^{\sqrt[]{10}} ~ i ~ di \\ \displaystyle = \dfrac{i^{1+1}}{1+1} ~\Big|^{\sqrt[]{10}}_0\\ \displaystyle = \dfrac{i^2}{2}~ \Big|^{\sqrt[]{10}}_0\\ \displaystyle = \dfrac{x^2}{2}~ \Big|^{\sqrt[]{10}}_0 ~~ \boxed{\text{Since i = x}} \\ \displaystyle = \dfrac{(\sqrt[]{10})^2}{2} - \dfrac{(0)^2}{2} \\ \displaystyle = \dfrac{10}{2} = \dfrac{5}{1} \\ \displaystyle \boxed{\therefore \int_0^{\sqrt[]{10}} ~ \sqrt[]{x\sqrt[]{x\sqrt[]{x\sqrt[]{x\ldots}}}} ~ dx = 5}$

√(x√(x√(x√(..√x) .)) ) = y, xy = y^2, x=y ∫ x dx=(1/2) x^2, So the answer is 5.