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Let $$I_n=\int_0^\infty e^{-x^n}\,\mathrm dx=\frac1n\Gamma(\tfrac1n).$$ Then, we have $$I_2=\frac1{2\sqrt2}\sqrt{2\pi}$$ $$I_4=\frac1{4}\sqrt{\sqrt{2\pi}}\sqrt{2\varpi}$$ where $\pi=2\int_0^1\frac{\mathrm dx}{\sqrt{1-x^2}}=\Gamma\left(\tfrac12\right)^2$ is the circle constant and $\varpi=2\int_0^1\frac{\mathrm dx}{\sqrt{1-x^4}}=\frac{\Gamma\left(\tfrac14\right)^2}{2\sqrt{2\pi}}$ is the lemniscate constant.

Do we have the following expresion for $I_6$?: $$c\sqrt{\sqrt{\sqrt{2\pi}}}\sqrt{\sqrt{2\varpi}}\sqrt{2s}$$ where $s=2\int_0^1\frac{\mathrm dx}{\sqrt{1-x^6}}=\frac{\Gamma\left(\tfrac16\right)\Gamma\left(\tfrac13\right)}{2\sqrt{3\pi}}$ and $c$ is an algebraic integer.

Bob Dobbs
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    If a result like that can be proven, this is probably how. – J.G. May 11 '25 at 13:13
  • @J.G.I asked for yes/no and a reference. I don't understand the downvotes. – Bob Dobbs May 11 '25 at 13:19
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    I don't understand the downvotes either but perhaps you can tell what $s$ is otherwise the question sounds vacuous. – lcv May 11 '25 at 13:21
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    Not only do we not know which curve you have in mind or how a constant is defined in terms of it, we don't know why that particular function of $s$ is conjectured. At this stage, the material I originally suggested is all I can offer you. I suggest you use it to rearrange your formula into an expression for $s$ you can try simplifying. Then it might be obvious how it's related to some curve. (By the way, I didn't downvote you.) – J.G. May 11 '25 at 13:24
  • I did not want to write a mathematically wrong equation. That is why I didn't write the expression for $s$. – Bob Dobbs May 11 '25 at 13:35
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    Here's a video by Michael Penn which deals with this exact problem: https://www.youtube.com/watch?v=Nn5WadFKAew – Supernerd411 May 11 '25 at 13:45
  • You can't possibly ask a question without writing a wrong equation. The whole point of asking the question is that you don't know whether the equations are wrong or not! – Trebor May 11 '25 at 13:57
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    Unless I somehow managed to make a mistake throwing your $I_4$ into Python, the expression is off by a factor of $\sqrt{2}$. The correct version is $I_4=\Gamma(1 + 1/4) = \frac{1}{4} \sqrt{\sqrt{2\pi}} \sqrt{2\omega} = 0.9064024770554769...$. Further, ignoring the $\frac{1}{2\sqrt{6}}$ factor, we have $\frac{\text{your }I_6}{\text{true }I_6} = 4.5234672653082875...$, which is unlikely to be a simple $\sqrt{2}$ fraction, because WolframAlpha doesn't recognize it. – Noctis May 11 '25 at 14:31
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    @Noctis Thanks. I will check $I_4$ again. I derived it myself. I should have done a mistake somewhere, if you are right. – Bob Dobbs May 11 '25 at 14:44
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    @Noctis How about now? I changed the expression for $I_6$ by linear logic. – Bob Dobbs May 11 '25 at 15:23
  • Your proposed expression for $I_6$ gives $0.7418448224\ldots$, whereas $\frac{1}{6}\Gamma(\frac{1}{6})=0.9277193332\ldots$. – Gary Jun 12 '25 at 00:54
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    @Gary Yes. Badly asked question. I edited. Thanks for numerical checking which I should have done much before. – Bob Dobbs Jun 12 '25 at 01:02

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We have $$\Gamma(\tfrac16)=\frac{3^{\frac12}\Gamma(\tfrac13)^2}{2^{\frac13}\pi^{\frac12}}.\tag1$$ Let $I=\sqrt{\sqrt{\sqrt{2\pi}}}\sqrt{\sqrt{2\varpi}}\sqrt{2s}$. Then, we have $\sqrt{2s}=\frac{\Gamma(\tfrac13)^{\frac32}}{2^{\frac16}\pi^\frac12}$ and $I=\frac{\Gamma(\tfrac14)^{\frac12}\Gamma(\tfrac13)^{\frac32}}{2^{\frac16}\pi^\frac12}.$

Supose that we have a formula like $I_6=\frac16\Gamma(\tfrac16)=cI$ where $c$ is an algebraic integer. Then, $$\frac{\Gamma(\tfrac13)}{\Gamma(\tfrac14)}=\sqrt[3]{128}c^2$$ must be an algebraic integer which is not likely.

Bob Dobbs
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