The title says it all. No trivial answers like $\int_0^\pi e^tdt$ please. The idea is rather, if there are integrals like $$\int\limits_0^\infty \frac{t^{2n}}{\cosh t}dt=(-1)^{n}\left(\frac{\pi}2\right)^{2n+1}E_{2n}$$and $$\int\limits_0^\infty \frac{t^{2n-1}e^{-t}}{\cosh t}dt=(-1)^{n-1}\frac{2^{2n-1}-1}{n}\left(\frac{\pi}2\right)^{2n}B_{2n}$$ (here, $E_{2n}$ and $B_{2n}$ are Euler and Bernoulli numbers), there should also be integrals of similar type that yield $e^\pi$ or $e^{-\pi}$. Certainly not by means of the given ones. Any ideas?
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2$\begingroup$ You better to show work about your question since we can't confirm your answer and it is harmful to yourself if we just do all the work for you $\endgroup$– VictorCommented Mar 20, 2012 at 23:24
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2$\begingroup$ One can always find something that works. I have gathered a dozen or so pages of integrals. And the closest I have come is integrals giving $e/\pi$, the closest "non trivial" to your question is the following. $$ \int_R \frac{1}{\sqrt{\pi}}e^{-x^2}\cos\left( 2\sqrt{\pi}x \right) \, \mathrm{d}x = e^{-\pi} \\ \int_R \frac{x \sin x}{1+x^2} \, \mathrm{d}x = \frac{e}{\pi}$$, where $R$ denote the whole numberline. (minus infinity to infinity) $\endgroup$– N3buchadnezzarCommented Mar 21, 2012 at 0:05
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$\begingroup$ OK, I see that the first one is $\int_R(e^{-x^2} \cos 2x) dx=\frac{\sqrt {\pi}}{2e}$ rescaled... The second one is very nice, although it should be $\frac{\pi}e$ instead of $\frac e{\pi}$. $\endgroup$– WolfgangCommented Mar 21, 2012 at 11:14
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$\begingroup$ Not an integral, but here's a fun probability puzzle with the answer $e^{\pi/4}$: youtube.com/watch?v=6_yU9eJ0NxA $\endgroup$– Hans LundmarkCommented Dec 17, 2019 at 14:37
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$\begingroup$ See also this one: math.stackexchange.com/questions/2959421/is-pi-e-a-period $\endgroup$– GEdgarCommented Feb 11 at 19:57
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2 Answers
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What about these?
$$\int_{0}^{1} \left(\frac{5}{2} \left((x - \sqrt{x^2 - 1})^{2i} + x^4\right) - 1\right) \, dx = e^{\pi}\tag{1}$$ $$\int_{0}^{1} \left( -\frac{5}{2\left( x - \sqrt{x^2 - 1}\right)^{2i}} - \frac{5x^4}{2} + 1 \right) \, dx = e^{-\pi}\tag{2}$$
answered Feb 11 at 19:06
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It's not clear to me exactly what you mean by "trivial" here. Anything that mentions $\pi$ explicitly, in endpoints or integrand? But the exponential function is OK? How about these? $$\int_0^1 \left(1 + \frac{4}{1+x^2} e^{4 \arctan(x)}\right)\ dx = e^\pi$$ $$\int_0^1 \left(1 - \frac{4}{1+x^2} e^{-4 \arctan(x)}\right)\ dx = e^{-\pi}$$
answered Mar 21, 2012 at 1:40
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1$\begingroup$ Also: $$\int_0^\infty \left(e^{-x} + \frac{2}{1+x^2} e^{2 \arctan(x)} \right)\ dx = e^\pi$$ $$\int_0^\infty \left(e^{-x} - \frac{2}{1+x^2} e^{-2 \arctan(x)} \right)\ dx = e^{-\pi}$$ $\endgroup$ Commented Mar 21, 2012 at 6:06