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While exploring possible applications for this new trick, I stumbled upon an entire family of integrals that "always" yield $a\pi^n$, where $a$ is an algebraic number and $n$ is a natural number. The following integral captivated me greatly:

$$\boxed{\int_{-1}^{1} \frac{125}{12}\sqrt[10]{\frac{1 + x}{1 - x}} (x^2 - x) \, dx = \color{red}{\phi}\color{blue}{\pi}}\tag{1}$$

The family:

$$\int_{-1}^{1} \left( \frac{1 + x}{1 - x} \right)^{\frac{1}{2}} (x^2 - x) \, dx = 0$$

$$\int_{-1}^{1} \left( \frac{1 + x}{1 - x} \right)^{\frac{1}{4}} (x^2 - x) \, dx = \frac{\pi}{8\sqrt{2}}$$

$$\int_{-1}^{1} \left( \frac{1 + x}{1 - x} \right)^{\frac{1}{6}} (x^2 - x) \, dx = \frac{10\pi}{81}$$

$$\int_{-1}^{1} \left(\frac{1 + x}{1 - x}\right)^{\frac{1}{8}} (x^2 - x) \, dx = \frac{\sqrt{3}}{100} \left(\frac{22787}{479}\right)^{\frac{1}{4}} \pi^2$$

$$\int_{-1}^{1} \left( \frac{1 + x}{1 - x} \right)^{\frac{1}{10}} (x^2 - x) \, dx = \frac{12}{125}\phi\pi$$

$$\int_{-1}^{1} \left( \frac{1 + x}{1 - x} \right)^{\frac{1}{12}} (x^2 - x) \, dx = \frac{15455288\pi}{94257441}$$

  1. Do you know of other cases of integrals that yield $\phi\pi$ in a non-obvious manner?

  2. The family seems to beg for a generalization. Is such a thing possible?

Emmanuel José García
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1 Answers1

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Let's concider $$I(\alpha)=\int_{-1}^1\left(\frac{1+x}{1-x}\right)^\alpha x(x-1)dx;\,\,-1<\alpha<2$$ Making the substitution $\frac{1+x}{1-x}=t$ $$I(\alpha)=4\int_0^\infty\frac{t^\alpha}{(1+t)^4}(1-t)dt$$ Making the substitution $x=\frac1{1+t}$ $$=4\int_0^1(1-x)^\alpha x^{2-\alpha}dx-4\int_0^1(1-x)^{\alpha+1}x^{1-\alpha}dx$$ Integrating the second term by parts $$=4\frac{1-2\alpha}{2-\alpha}\int_0^1(1-x)^\alpha x^{2-\alpha}dx=4\frac{1-2\alpha}{2-\alpha}B\big(1+\alpha;3-\alpha\big)=\frac23\frac{1-2\alpha}{2-\alpha}\Gamma(1+\alpha)\Gamma(3-\alpha)$$ Using the Euler formula for Gamma-function $$I(\alpha)=\frac23\frac{1-2\alpha}{2-\alpha}\alpha(2-\alpha)(1-\alpha)\Gamma(\alpha)\Gamma(1-\alpha)=\frac{2\pi}3\frac{\alpha(1-\alpha)(1-2\alpha)}{\sin\pi\alpha}$$ You can quickly check the answer, for example, at $\alpha=\frac16$ and get $\displaystyle\frac{10\pi}{3^4}$

Svyatoslav
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    This is elegant and beautiful. Thanks for posting this kind of answers. Cheers :-) – Claude Leibovici Mar 28 '24 at 08:36
  • @Claude Leibovici , thank you for your kind words. Have a nice day! – Svyatoslav Mar 28 '24 at 08:41
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    (+1) Wow! Due to this, we can also see that when $\alpha$ is rational, the answer is $\pi$ times some algebraic number. But this also means that the answer in the original post is wrong (I manually checked that the answer given by the formula in this post does not match). – ultralegend5385 Mar 28 '24 at 12:43
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    @ultralegend5385, yes, the answer is in general $\pi$ times some algebraic number. I have some doubt about the answer at $\alpha=\frac18$, but the answers at $\alpha=\frac12, ,\frac14,,\frac16$ are correct – Svyatoslav Mar 28 '24 at 12:56
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    You are correct about my integral giving $\phi\pi$. That's incorrect. But the difference is 0.000000000000384581. I'm sorry, it was 2:00 a.m. and I got these results empirically. So the original integral is just almost $\phi\pi$. – Emmanuel José García Mar 28 '24 at 16:04
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    I guess your answer is however correct, given that $\sin\frac\pi{10}=\frac{\sqrt 5-1}4$ :) https://www.wolframalpha.com/input?i=%5Csin%28%5Cpi%2F10%29 – Svyatoslav Mar 28 '24 at 16:37