Algebraic or Not

$\begin{aligned} I & = \int_0^\infty \frac 1{1+x^{16}} dx & \small \blue{\text{Let }\tan \theta = x^8 \implies \sec^2 \theta\ d\theta = 8x^7\ dx} \\ & = \int_0^\frac \pi 2 \frac 1{8 \tan^\frac 78 \theta} d\theta \\ & = \frac 18 \int_0^\frac \pi 2 \sin^{-\frac 78} \theta \cos^\frac 78 \theta \ d \theta & \small \blue{\text{Beta function }\text B(m,n) = 2 \int_0^\infty \sin^{2m-1} (t) \cos^{2n-1}(t)\ dt} \\ & = \frac 1{16} \text B\left(\frac 1{16}, \frac {15}{16}\right) & \small \blue{\text{and B}(m.n) = \frac {\Gamma (m)\Gamma(n)}{\Gamma(m+n)}, \Gamma (\cdot) \text{ is gamma function.}} \\ & = \frac {\Gamma \left(\frac 1{16}\right)\Gamma\left(\frac {15}{16}\right)}{16\Gamma (1)} & \small \blue{\text{Note that }\Gamma(s) \Gamma(1-s) = \frac \pi{\sin \pi s}} \\ & = \frac \pi {16 \sin \frac \pi{16}} \end{aligned}$

Therefore $\dfrac I\pi = \dfrac 1{16} \csc \dfrac \pi{16}$ , which is an algebraic number if $\csc \dfrac \pi{16}$ is an algebraic number. Consider the following:

$\begin{aligned} \cos \frac \pi 4 & = \frac 1{\sqrt 2} \\ 2\cos^2 \frac \pi 8 - 1 & = \frac 1{\sqrt 2} \\ 2\left(1-2\sin^2 \frac \pi{16}\right)^2 - 1 & = \frac 1{\sqrt 2} & \small \blue{\text{Let }\csc \frac \pi{16}=x} \\ 2\left(1- \frac 2{x^2} \right)^2 - 1 & = \frac 1{\sqrt 2} \\ \frac {2(x^2-2)^4}{x^4} - 1 & = \frac 1{\sqrt 2} & \small \blue{\text{Multiply both sides by }x^4} \\ 2(x^4-4x^2+4) - x^4 & = \frac {x^4}{\sqrt 2} \\ x^4-8x^2+8 & = \frac {x^4}{\sqrt 2} & \small \blue{\text{Squaring both sides}} \\ x^8-16x^6 + 80x^4 -128x^2 + 64 & = \frac {x^8}2 & \small \blue{\text{Rearranging}} \\ x^8 -32x^6 + 160x^4 -256x^2 + 128 & = 0 & \small \blue{\text{Minimal polynomial}} \end{aligned}$

Yes , $\dfrac I\pi$ is an algebraic number since $\csc \dfrac \pi {16}$ is an algebraic number.

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References:

• Algebraic number
• Beta function
• Gamma function

The integral is $\dfrac{1}{16\sin(\pi/16)}$ , which comes from a standard application of contour integration.

Now, $\dfrac{1}{\sqrt{2}}=\cos(\pi/4)=2\cos^{2}(\pi/8)-1=2(1-2\sin^{2}(\pi/16))^{2}-1$ and by squaring this expression, we get that $\sin(\pi/16)$ is algebraic.

Then, one can prove easily that an algebraic number divided by an integer is also algebraic. And $\dfrac{1}{\alpha}$ is algebraic if $\alpha$ is algebraic and non zero. And so, the result follows.