Is there a non-trivial definite integral that values to $\frac{e}{\pi}$?

I know that both $$\int_{-\infty}^\infty \frac{\cos(x)}{x^2+1} \, \text{dx} = \frac{\pi}{e}$$ and $$\int_{-\infty}^\infty \frac{x\sin(x)}{x^2+1} = \frac{\pi}{e}$$ But what about $\frac{e}{\pi}$? Is there a non-trivial definite integral which evaluates to that, but still where the integrand is composed of elementary functions like the two above? The only constraint on the integrand is that it is an elementary function. It does not need to have an elementary antiderivative. Also, the value of the integral should not be obvious at first glance. The integrals mentioned above are certainly not, which makes the answer so beautiful in my opinion.

By trivial I mean integrals like $\int_{-\infty}^1 \frac{e^x}{\pi} \, \text{d}x = \frac{e}{\pi}$, which quite obviously converges to the desired value.

Edit: I have been asked to clarify my question as it is too broad. The two integrals mentioned above are, in my opinion, beautiful because they both converges to a fraction containing two such fundamental constants, even though it is not obvious at all that they should at first glance. Then, out of curiosity, I wondered if there are similar integrals which converges to $\frac{e}{\pi}$. The answer @aleden has written is certainly interesting, but some of the magic disappears when both $\pi$ and $e$ are in the integrand. I have added some more information in the paragraphs above. I hope this cleared my question a bit, but it is broad by its nature.

By Frullani's Theorem:

$$\frac{1}{f(0)-f(\infty)}\int_0^\infty \frac{f(e^{\pi} x)-f(e^e x)}{x}dx=\frac{e}{\pi}$$ for appropriate functions $f(x)$.

EDIT: Alternatively, you can look for a function $f(x)$ that satisfies $\lim_{x\to\infty}f(x)=\frac{e}{\pi}$ such as $f(x)=\frac{ex}{\pi x+1}$ and use:

$$\frac{1}{\ln(\frac{a}{b})}\int_0^\infty \frac{eax(\pi bx+1)-ebx(\pi ax+1)}{x(\pi ax+1)(\pi bx+1)}dx=\frac{e}{\pi}$$