A Hard Looking Easy Problem From MIT

image containing math problem statement
A problem that looks intimidating at a glance but may not be very hard if you look closer. A problem from the MIT Integration Bee.
Problem contributed by debroy0611
As a first step, let us simplify the expression inside the integration sign:

$\sqrt{x \sqrt[3]{x \sqrt[4]{x \sqrt[5]{x...}}}}$

$= \sqrt{x} \cdot \sqrt{ \sqrt[3]{x \sqrt[4]{x \sqrt[5]{x...}}}}$

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$= x^{\frac{1}{2}} \cdot \sqrt{ \sqrt[3]{x \sqrt[4]{x \sqrt[5]{x...}}}}$

$= x^{\frac{1}{2}} \cdot \sqrt{ \sqrt[3]{x}} \sqrt{\sqrt[3]{\sqrt[4]{x \sqrt[5]{x...}}}}$

$= x^{\frac{1}{2}} \cdot x^{\frac{1}{2} \cdot \frac{1}{3}} \sqrt{\sqrt[3]{\sqrt[4]{x \sqrt[5]{x...}}}}$

$= x^{\frac{1}{2}} \cdot x^{\frac{1}{2} \cdot \frac{1}{3}} \sqrt{\sqrt[3]{\sqrt[4]{x}}} \sqrt{\sqrt[3]{\sqrt[4]{\sqrt[5]{x...}}}}$

$= x^{\frac{1}{2}} \cdot x^{\frac{1}{2} \cdot \frac{1}{3}} \cdot x^{\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4}} \cdot x^{\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{5}}...$

$= x^{\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{5}...}$

$= x^{\frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}...}$

The expression in the exponent is related to $e$, in the sense that:
$e = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dfrac{1}{5!}...$

$\Rightarrow e = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dfrac{1}{5!}...$

$\Rightarrow e - 2 = \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dfrac{1}{5!}...$

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Therefore, we get the integral as:
$\displaystyle \int x^{e-2} dx$

$= \dfrac{x^{e-1}}{e-1}$