Nested Square Roots within Integral

Let's "pull out" all the $x$ 's:

$= x^{\left(1\sqrt{2 \sqrt{3 \sqrt{...}}}\right)}$

$= x^{\left(\frac{1}{2^0} + \frac{2}{2^1} + \frac{3}{2^2} + ...\right)}$

Now, let's concentrate on finding the sum of the series. The sum can be represented in a form as follows:

$\frac{1}{2^0} + \frac{2}{2^1} + \frac{3}{2^2} + ... + \frac{2}{2^1} + \frac{3}{2^2} + ... + \frac{3}{2^2}+...$

Now, the sequence is represented as a sum of infinite geometric series. Summing them up, using the rule that $S_\infty = \frac{a}{1-r}$ :

$= 2 + 1 + \frac{1}{2} + ...$

$= 4$ .

Now, this is also an infinite geometric series, and its sum is 4. Substituting that value back into the original question:

$=\displaystyle \int_0^1 x^4 dx$

$= \frac{1}{5} x^5 \Big|_0^1$

$= \boxed{0.2}$

Note that $\displaystyle x\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5 \cdots}}}} = x \left(x^2 \left(x^3 \left(x^4 \left(x^5 (\cdots) \right)^\frac 12\right)^\frac 12 \right)^\frac 12 \right)^\frac 12 = x \cdot x^\frac 22 \cdot x^\frac 34 \cdot x^\frac 48 \cdot x^\frac 5{16} \cdots = x^{1+\frac 22 + \frac 34 + \frac 48 + \frac 5{16}+\cdots} = x^4$ (see note).

Therefore, $\displaystyle \int_0^1 x\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5 \cdots}}}} dx = \int_0^1 x^4 \ dx = \frac {x^5}5 \bigg|_0^1 = \frac 15 = \boxed{0.2}$ .

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Note: For $|x| < 1$ ,

$\begin{aligned} \frac 1{1-x} & = 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots & \small \blue{\text{Differentiate both sides w.r.t }x} \\ \frac 1{(1-x)^2} & = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \cdots & \small \blue{\text{Replace }x \text{ with }\frac 12} \\ 4 & = 1+\frac 22 + \frac 34 + \frac 48 + \frac 5{16}+\cdots \end{aligned}$