MHB Is the Integral of 1/sqrt(1-x^4) Expressible in Terms of Elementary Functions?

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The integral of 1/sqrt(1-x^4) cannot be expressed in terms of elementary functions. Evidence presented includes references to Chebyshev's theorem, which confirms that the specific parameters of the integral do not meet the criteria for expressibility in elementary terms. The discussion notes that while tools like Mathematica may not fully implement the Risch algorithm, the conclusion remains valid. The parameters of the integral indicate that none of the necessary conditions for expressibility are satisfied. Therefore, the integral is confirmed to require special functions for its representation.
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$\displaystyle (3)\;\; \int \frac{1}{\sqrt{1-x^4}}dx$
 
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jacks said:
$\displaystyle (3)\;\; \int \frac{1}{\sqrt{1-x^4}}dx$

Hi jacks, :)

This integral cannot be expressed in terms of elementary functions. See this.

Kind Regards,
Sudharaka.
 
Sudharaka said:
Hi jacks, :)

This integral cannot be expressed in terms of elementary functions. See this.

Kind Regards,
Sudharaka.

That is strong evidence, but is not water-tight as IIRC neither Mathematica nor Alpha implements the full Risch algorithm and will occasionally drop through to special functions where an elementary integral does exist.

CB
 
Chebyshev's theorem: If $a, b \in\mathbb{R}$ and $m,p,n \in\mathbb{Q}$ then the (indefinite) integral of $ x^m\left(a+bx^n\right)^p$ can be written in terms of elementary functions if and only if one of $p,~ (m+1)/n, ~p+ (m+1)/n ~\in\mathbb{Z}$. In our case we have $m=0, ~ p = -\frac{1}{2}, ~ a = 1, ~ b = -1$ and $n = 4$. Clearly $p = -\frac{1}{2} \not\in\mathbb{Z}, ~ (m+1)/n = \frac{1}{4} \not\in\mathbb{Z}$ and $p+(m+1)/n = -\frac{1}{4} \not\in\mathbb{Z}.$ Thus $\int \frac{1}{\sqrt{1-x^4}}\;{dx}$ cannot be written in terms of elementary functions.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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