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

Question

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?

Answer 1

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? $$\int_0^1 \left(1 + \frac{4}{1+x^2} e^{4 \arctan(x)}\right)\ dx = e^\pi$$ $$\int_0^1 \left(1 - \frac{4}{1+x^2} e^{-4 \arctan(x)}\right)\ dx = e^{-\pi}$$