Is there a non-trivial integral yielding $e$?
Similar questions have been asked here and here. However, both of these posts asks for integrals yielding elementary functions of $e$, and not necessarily $e$ itself. Their accepted answers are $$\int_0^\infty\cos\left(x\sqrt{x^2+b^2}\right)dx = \frac{\sqrt{2\pi}}4 e^{-b^2/2}$$ and a particular case of $$\int_0^\infty\frac{\cos(bx)}{a^2+x^2}dx = \frac\pi {2a} e^{-|ab|}$$
I do not take issue with the coefficients multiplying $e$, but with the fact its exponent is necessarily negative in both cases. Can we find a non-trivial integral yielding $e$ itself? Of course, I'm not asking about the existence of such an integral (one certainly exists) but for an explicit example not relying on $e$ itself on its input.
One of the previous posts mentions it is an open problem to determine if $e$ is a period and it is believe not to be, so this reduces the scope of possibilities for integrals yielding $e$.
On a similar note, this post notices there are simple and beautiful integral representations for $\dfrac\pi e$ and asks for an integral yielding $\dfrac e\pi$. It remains unanswered with respect to the criteria above.
Here is another interesting integral, unfortunately relying on $e$ in its input $$\int_1^\infty x^{-x} e^x\ln x~dx = e$$
As mentioned in the comments, this has also been asked in this post and received the answer $$e = \int_0^\infty\frac1{\lfloor x\rfloor!}dx$$ This is nothing more than a sketchy way of writing $\displaystyle\sum_{n=0}^\infty\frac1{n!}$ and every series have a similar integral form, so I'd say this is one trivial representation.
Another comment asks about the strategy of milking one trivial integral until it becomes non-trivial. I'm sure this can result in interesting identities, however, the point of this post in particular is a desire to develop a further understanding of the constant $e$ and its properties. In this sense, reverse engineering expressions leaves a lot to be desired. I don't think I can really rule out "reverse engineering" as a strategy, as any identity can be seen as a reverse engineered identity, but I hope this comment elucidate my issue with some of the expressions suggested.
The user Tyma Gaidash commented below one very simple, and yet extremely clever, approach to this problem, which is to integrate the derivative of the function $\displaystyle\left(1+\frac1x\right)^x$. We have $$\frac d{dx}\left(1+\frac1x\right)^x = \left(1+\frac1x\right)^x\left(\ln\left(1+\frac1{x}\right)-\frac1{x+1}\right)$$ therefore $$\int_0^\infty \left(1+\frac1x\right)^x\left(\ln\left(1+\frac1{x}\right)-\frac1{x+1}\right)dx = \left.\left(1+\frac1x\right)^x\right|_0^\infty = e-1$$ This is very close to what I'm looking for, the only issue being the presence of the logarithm.
