# 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 $$\int\limits_0^\infty \frac{t^{2n}}{\cosh t}dt=(-1)^{n}\left(\frac{\pi}2\right)^{2n+1}E_{2n}$$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?

• 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 Commented Mar 20, 2012 at 23:24
• 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) Commented Mar 21, 2012 at 0:05
• 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 $\frac e{\pi}$. Commented Mar 21, 2012 at 11:14
• Not an integral, but here's a fun probability puzzle with the answer $e^{\pi/4}$: youtube.com/watch?v=6_yU9eJ0NxA Commented Dec 17, 2019 at 14:37
• See also this one: math.stackexchange.com/questions/2959421/is-pi-e-a-period Commented Feb 11 at 19:57

$$\int_{0}^{1} \left(\frac{5}{2} \left((x - \sqrt{x^2 - 1})^{2i} + x^4\right) - 1\right) \, dx = e^{\pi}\tag{1}$$ $$\int_{0}^{1} \left( -\frac{5}{2\left( x - \sqrt{x^2 - 1}\right)^{2i}} - \frac{5x^4}{2} + 1 \right) \, dx = e^{-\pi}\tag{2}$$
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? $$\int_0^1 \left(1 + \frac{4}{1+x^2} e^{4 \arctan(x)}\right)\ dx = e^\pi$$ $$\int_0^1 \left(1 - \frac{4}{1+x^2} e^{-4 \arctan(x)}\right)\ dx = e^{-\pi}$$
• Also: $$\int_0^\infty \left(e^{-x} + \frac{2}{1+x^2} e^{2 \arctan(x)} \right)\ dx = e^\pi$$ $$\int_0^\infty \left(e^{-x} - \frac{2}{1+x^2} e^{-2 \arctan(x)} \right)\ dx = e^{-\pi}$$ Commented Mar 21, 2012 at 6:06