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?

  • 2
    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 – Victor Mar 20 '12 at 23:24
  • 2
    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) – N3buchadnezzar Mar 21 '12 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 $\frac e{\pi}$. – Wolfgang Mar 21 '12 at 11:14
up vote 4 down vote accepted

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: $$\int_0^\infty \left(e^{-x} + \frac{2}{1+x^2} e^{2 \arctan(x)} \right)\ dx = e^\pi$$ $$\int_0^\infty \left(e^{-x} - \frac{2}{1+x^2} e^{-2 \arctan(x)} \right)\ dx = e^{-\pi}$$ – Robert Israel Mar 21 '12 at 6:06
  • Nice examples. Thank you! – Wolfgang Mar 21 '12 at 8:19
  • @RObert They are good examples, but a change of variables makes them trivial. – Pedro Tamaroff Mar 23 '12 at 2:21

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.