Integral #5

A good way to solve this problem is to use partial sums and then extrapolate the series.

• $\sqrt{x} = x^{\frac{1}{2}}$

• $\sqrt{x \sqrt[3]{x}} = x^{\frac{\frac{1}{3} + 1}{2}} = x^{\frac{1}{6} + \frac{1}{2}} = x^{\frac{2}{3}}$

• $\sqrt{x \sqrt[3]{x \sqrt[4]{x}}} = x^{\frac{\frac{\frac{1}{4} + 1}{3} + 1}{2}} = x^{\frac{1}{24} + \frac{1}{6} + \frac{1}{2}} = x^{\frac{17}{24}}$

• $\sqrt{x \sqrt[3]{x \sqrt[4]{x \sqrt[5]{x}}}} = x^{\frac{1}{120} + \frac{1}{24} + \frac{1}{6} + \frac{1}{2}} = x^{\frac{43}{60}}$

If this process was repeated infinitely many times, it appears that the radicals would equal $x^{k}$ where $k = \sum_{2}^{\infty} \frac{1}{n!} = \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ...$

Recall that the Maclaurin series for $e^{x}$ is $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...$ . Thus, $e - 2 = \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ...$

The integral now becomes $\int_0^e x^{e - 2} dx$ . Solving yields $\frac{1}{e - 1}x^{e-1} \rvert^e_0 = \frac{1}{e - 1}e^{e - 1}$ .

Therefore $a = \frac{1}{e - 1}, b = e, c = e - 1$ and $a + b + c = \boxed{5.01854}$

The integrand is $x^{\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...}=x^{e-2}$ . So the value of the integral is $\dfrac{e^{e-1}}{e-1}$ . Hence $a=\dfrac{1}{e-1}, b=e, c=e-1$ and $a+b+c=\dfrac{2e^2-3e+2}{e-1}\approx \boxed {5.01854036379}$

$\begin{aligned} I & =\int_0^e \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]\cdots}}} \ dx \\ & = \int_0^e \left(x \left(x \left(x \left(\cdots \right)^\frac 15\right)^\frac 14\right)^\frac 13\right)^\frac 12 dx \\ & = \int_0^e x^{\frac 1{2!}+\frac 1{3!}+\frac 1{4!} + \cdots} dx \\ & = \int_0^e x^{e-2} dx \\ & = \frac {x^{e-1}}{e-1} \bigg|_0^e \\ & = \frac {e^{e-1}}{e-1} \end{aligned}$

Therefore $a+b+c = \dfrac 1{e-1} + e + e-1 \approx \boxed{5.019}$ .