I gave this integral as an answer at MathSE.

\[\int_{0}^{1}\left(\frac{5}{2} \left((x - \sqrt{x^2 - 1})^{2i} + x^4\right)-1\right)\,dx=e^\pi.\]

The integral yields \(e^{\pi}\). I don't know why the LaTeX isn't displaying properly.

Link: https://math.stackexchange.com/questions/122693/is-there-a-definite-integral-that-yields-e-pi-or-e-pi-in-a-non-trivial/4861138?fbclid=IwAR3CPXtBoNDY3yc5a1HDAgfWN7QFAO2Ph91uvH7MUgSxSRjsq-t8TWYHWdY#4861138

#math #integral #eulernumber #halfangleapproach #symmetrymatters #calculus

Is there a definite integral that yields $e^\pi$ or $e^{-\pi}$ in a non trivial way?

The title says it all. No trivial answers like $\int_0^\pi e^tdt$ please. The idea is rather, if there are integrals like $$\int\limits_0^\infty \frac{t^{2n}}{\cosh t}dt=(-1)^{n}\left(\frac{\pi}2\r...

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Deleting #ephipi cruft. The first #Perl script I'm sticking with (which I've renamed ePhiPi_0xMagBin.pl) is the poor man's stress test. All it does it examine the constants #EulerNumber, #Phi, and #pi for matching single binary digits or strings of digits by orders of 0x magnitude. It reaches the 7th fairly quickly.

Is \(\left\lfloor\dfrac{n!}{e}\right\rfloor,\ \forall n\in\mathbb{N}\) always even? or equivalently, is \(\dfrac{1}{2}\left\lfloor\dfrac{n!}{e}\right\rfloor,\ \forall n\in\mathbb{N}\) always an integer?🤔 🔗 https://www.youtube.com/watch?v=wrHxeHJDTk4&t=604s&ab_channel=MichaelPenn

#floorfunction #function #even #EulerNumber #Factorial #NaturalNumber #MichaelPenn

Always even.

YouTube