Root in Root Integral

Calculus Level 3

Behold the indefinite integral: $\int_{ }^{ }\left(\frac{\sqrt[2]{x^{\sqrt{3}}}\sqrt[4]{x^{\sqrt{3}}}\sqrt[14]{x^{\sqrt{3}}}\sqrt[16]{x^{\sqrt{3}}}...}{\sqrt[8]{x^{\sqrt{3}}}\sqrt[10]{x^{\sqrt{3}}}\sqrt[20]{x^{\sqrt{3}}}\sqrt[22]{x^{\sqrt{3}}}...}\right)dx$ Hint: The Integrand can be algebraically manipulated to show a x taken to the exponent of a infinite series: $\int_{ }^{ }x^{\sum_{n=1}^{\infty}\frac{\sin\left(n\theta\right)}{n}}dx$

Answer Submission: The integral can be expressed as $\frac{x^{\left(\alpha\pi+1\right)}}{\alpha\pi+1}+C$ , the answer is $\alpha^{-2}$ .

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The answer is 9.

1 solution

Tom Engelsman
May 2, 2020

The integrand's numerator and denominator can be respectively written as:

$(x^{\sqrt{3}})(\frac{1}{2}+\frac{1}{4} +\frac{1}{14}+\frac{1}{16}+ … + \frac{1}{12n-10} + \frac{1}{12n-8} + ... )$ (i)

$(x^{\sqrt{3}})(\frac{1}{8}+\frac{1}{10} +\frac{1}{20}+\frac{1}{22}+ … + \frac{1}{12n-4} + \frac{1}{12n-2} + ... )$ (ii)

Thus the integrand becomes:

$x^{\sqrt{3} \cdot N}$ , where $N = \Sigma_{k=1}^{\infty} \frac{1}{12k-10} + \frac{1}{12k-8} - \frac{1}{12n-4} - \frac{1}{12n-2} = \frac{\pi}{3\sqrt{3}}.$

or $\int x^{\pi / 3} dx = \boxed{\frac{x^{\pi/3 + 1}}{\pi/3 + 1} + C}$

which yields $\alpha = \frac{1}{3} \Rightarrow \alpha^{-2} = \boxed{9}.$

Root in Root Integral

The answer is 9.

1 solution

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