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Mathematica claims that $${\large\int}_0^1\!\!\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac{\sqrt\pi}{2^{\sqrt2}\sqrt2}\cdot\frac{\Gamma\left(\frac1{\sqrt2}\right)}{\Gamma\left(\frac12+\frac1{\sqrt2}\right)},\tag{$\diamond$}$$ and it also confirms numerically.

How can we prove $(\diamond)$?

This result seems interesting, because no such nice answer seems to occur for other algebraic powers, except trivial cases, when the antiderivative is an elementary function, e.g. $${\large\int}\frac{dx}{(1+x^\alpha)^\alpha}=\\ \small\frac{x\left(x^\alpha+1\right)^{1-\alpha}}{18}\left[\vphantom{\large|}\left(15(\alpha-1)+4\left(5(\alpha+1)+\left(6\alpha+(\alpha-3)x^{2\alpha}+2(\alpha+2)x^\alpha-3\right)x^\alpha\right)x^\alpha\right)x^\alpha+18\right],$$ where $\alpha=3+\sqrt{10}$.

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  • $\begingroup$ Looks a lot like Beta function to me at the first glance $\endgroup$ – Yuriy S Aug 16 '16 at 21:06
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Using Euler's integral representation of the Gaussian hypergeometric function, and assuming $a>0$, we get

$$ \begin{align} \int_{0}^{1} \frac{dx}{(1+x^{a})^{a}} &= \frac{1}{a} \int_{0}^{1} \frac{u^{1/a-1}}{(1+u)^{a}} \, du \\ &= \frac{1}{a} \int_{0}^{1} u^{1/a-1} (1-u)^{(1+1/a)-1/a-1}(1+u)^{-a} \, du \\ &= \frac{1}{a} \, B \left(\frac{1}{a}, 1 \right) {}_2F_{1} \left(a, \frac{1}{a}; 1+ \frac{1}{a}; -1 \right) \\&= {}_2F_{1} \left(a, \frac{1}{a}; 1+ \frac{1}{a}; -1 \right). \end{align}$$

In the case $a=\sqrt{2}$, we can apply Kummer's theorem since $1+\sqrt{2} - \frac{1}{\sqrt{2}}= 1+ \frac{1}{\sqrt{2}}$.

$$\begin{align} \int_{0}^{1} \frac{dx}{(1+x^{\sqrt{2}})^{\sqrt{2}}} &= \frac{\Gamma\left(1+ \frac{1}{\sqrt{2}}\right) \Gamma\left(1+\frac{1}{\sqrt{2}}\right)}{\Gamma\left(1+ \sqrt{2}\right) \Gamma\left(1\right)} \\ &= \frac{\Gamma \left(1+ \frac{1}{\sqrt{2}} \right)^{2}}{\Gamma(1+\sqrt{2})}\\ &= \frac{1}{2 \sqrt{2}} \frac{\Gamma\left(\frac{1}{\sqrt{2}}\right)^{2}}{\Gamma(\sqrt{2})} \\ &\approx 0.6604707226 \end{align}$$

To get it in the form given by Mathematica, apply the duplication formula to $\Gamma \left(\frac{1}{\sqrt{2}} \right)$.

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Hint. One may observe that $$ \begin{align} \int_0^\infty\frac{dx}{(1+x^a)^a}&=\int_0^1\frac{dx}{(1+x^a)^a}+\int_1^\infty\frac{dx}{(1+x^a)^a} \\\\&=\int_0^1\frac{dx}{(1+x^a)^a}+\int_0^1\frac{dx}{\left(1+\frac1{x^a}\right)^a\:x^2} \quad (x \to 1/x) \\\\&=\int_0^1\frac{dx}{(1+x^a)^a}+\int_0^1\frac{x^{a^2-2}dx}{(1+x^a)^a} \\\\&=\int_0^1\frac{1+x^{a^2-2}}{(1+x^a)^a}\:dx \end{align} $$ giving, for $a:=\sqrt{2}$, that $$ \int_0^1\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac12\int_0^\infty\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac{\Gamma\left(1+\frac{1}{\sqrt2}\right) \Gamma\left({\sqrt2}-\frac{1}{\sqrt2}\right)}{2\:\Gamma\left(\sqrt2\right)} $$ by using the Euler beta result with the change of variable $t=\dfrac1{(1+x^{\sqrt2})^{\sqrt2}}$ in the latter integral.

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    $\begingroup$ Neat the way this shows why $\sqrt{2}$ is special. I noticed that is $a^2-2 = a$, another interesting result might appear. $\endgroup$ – marty cohen Aug 16 '16 at 21:20

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