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I'm trying to prove $$I := \int_{0}^{\pi/2}{\sqrt{1 + \sqrt{1 + (\tan{x})^{2/3}}}\,dx} = \frac{\pi}{2} (3^{1/4} + 3^{3/4} - 2)$$ which is around $2.5063328837$.

Using the $u$-substitution $u = \sqrt{-1 + \sqrt{1 + (\tan{x})^{2/3}}}$, we have \begin{align*} du &= \frac{1}{6}\frac{1}{\sqrt{1 + (\tan{x})^{2/3}}}\frac{1}{\sqrt{-1 + \sqrt{1 + (\tan{x})^{2/3}}}} (\tan{x})^{-1/3} \sec^2{x}\,dx \\ &= \frac{1}{6 u (u^2 + 1)} \frac{1}{\sqrt{(u^2 + 1)^2 - 1}} \left(((u^2 + 1)^2 - 1)^3 + 1\right) \,dx \\ &= \frac{1}{6 u^2 (u^2 + 1)} \frac{1}{\sqrt{u^2 + 2}} \left(u^6 (u^2 + 2)^3 + 1\right) \,dx \end{align*} and thus, $$I = \int_{0}^{\pi/2}{\sqrt{1 + \sqrt{1 + (\tan{x})^{2/3}}}\,dx} = 6 \int_{0}^{\infty}{\frac{u^2 (u^2 + 1) (u^2 + 2)}{u^6 (u^2 + 2)^3 + 1} \,du}$$ So we have the integral of a rational function, where the roots of the denominator can easily be found in closed form; hence using the method of residues, we can get an answer in closed form. Yet, I've done this and it's still not so obvious that the answer miraculously simplifies to $\frac{\pi}{2} (3^{1/4} + 3^{3/4} - 2)$.

My question is, seeing that the answer is so nice, is there a more clever way to solve this integral? I've tried different substitutions, integration by parts, the Feynman trick by putting a parameter in place of the exponent $2/3$, etc., but they all seem to go nowhere.

Thanks!

Ant
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    What do you mean by "not obvious" do you manage with residues or do you just want a different way? – Diger Jun 19 '20 at 22:07
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    If you let $u=x^4+2x^2$, you get $\frac{3}{2} \int_0^{\infty} \frac{u}{x\left(u^3+1\right)} ; du$. That pesky x is being annoying. – Ty. Jun 19 '20 at 22:27

1 Answers1

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The explanation for simplicity of the result follows the same reasoning as in my answer here.

You obtained $$\frac{I}{6} = \int_{0}^{\infty}{\frac{u^2 (u^2 + 1) (u^2 + 2)}{u^6 (u^2 + 2)^3 + 1} \,du}$$ The denominator factors over $\mathbb{Q}(i\sqrt[4]{3})$ as $$u^6 (u^2 + 2)^3 + 1 = (u^2+1)^2 h(x) \overline{h}(x)$$ where $$h(x) = u^4+\sqrt[4]{3} i u^3+3^{3/4} i u^3-2 \sqrt{3} u^2-u^2-2 i \sqrt[4]{3} u+1$$ Note that $h$ has a very special distribution of zeroes: all of its zeroes lie in lower half plane (this has to checked numerically). Let $p(x) = u^2 (u^2+2)$, then residue theorem says $$\frac{I}{{12\pi i}} = - \sum\limits_{\alpha :h(\alpha ) = 0} {\frac{{p(\alpha )}}{{h'(\alpha )\overline{h}(\alpha )({\alpha ^2} + 1)}}} - \frac{{p( - i)}}{{( - 2i)h( - i)\overline{h}( - i)}} $$

First term of RHS is symmetric in roots of $h(x) \in \mathbb{Q}(i\sqrt[4]{3})[x]$, so it is in $\mathbb{Q}(i\sqrt[4]{3})$. Second term is in $\mathbb{Q}(i,i\sqrt[4]{3})$. Therefore $$I/\pi \in \mathbb{Q}(i,i\sqrt[4]{3}) \cap \mathbb{R} = \mathbb{Q}(\sqrt[4]{3})$$

Ant
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pisco
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