Another crazy integral

Calculus Level 4

Find the value of integral

∫ 0 1 ⁣ x x x x x 5 ⋯ 4 3 2 d x \large \int\limits_{0}^{1} \! x \sqrt[2]{x \sqrt[3]{x \sqrt[4]{x \sqrt[5]{x} \cdots}}}\, \mathrm{d} x


The answer is 0.3678.

Rajyawardhan Singh by Rajyawardhan Singh , 19, India

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2 solutions

I = ∫ 0 1 x x x x ⋯ 5 4 3 d x = ∫ 0 1 x ( x ( x ( x ( x ( x ⋯ ) ) 1 5 ) 1 4 ) 1 3 ) 1 2 d x = ∫ 0 1 x 1 1 ! + 1 2 ! + 1 3 ! + ⋯ d x = ∫ 0 1 x e − 1 d x = x e e ∣ 0 1 = 1 e ≈ 0.368 \large \begin{aligned} I & = \int_0^1 x\sqrt{x \sqrt[3] {x\sqrt[4] {x\sqrt[5] \cdots}}}\ dx \\ & = \int_0^1 x \left(x \left(x \left(x \left(x \left(x \cdots \right)\right)^\frac 15 \right)^\frac 14 \right)^\frac 13 \right)^\frac 12 dx \\ & = \int_0^1 x^{\frac 1{1!} + \frac 1{2!} + \frac 1{3!} + \cdots}\ dx \\ & = \int_0^1 x^{e-1}\ dx \\ & = \frac {x^e}e \ \bigg|_0^1 = \frac 1e \approx \boxed{0.368} \end{aligned}

3 Helpful 0 Interesting 3 Brilliant 0 Confused

Can we intigrate x^e as e is a transcendal number?

Alapan Das - 2 years, 1 month ago

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Yes, we can.

Chew-Seong Cheong - 2 years, 1 month ago

Small mistake on the result of the integral, should be e e in the exponent instead of e − 1 e-1

Guilherme Niedu - 2 years, 1 month ago

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Thanks, I will amend it.

Chew-Seong Cheong - 2 years, 1 month ago

The integrand is x^(e-1), so that the value of the integral is 1/e or 0.367879

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Is this taken from an MIT integration bee?

Zhang Xiaokang - 2 years, 1 month ago

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