Integral definition of e I know that $e$ can be defined via a convergent series: $$ e = \sum_{n=0}^\infty {1\over n!}$$
Or as a limit: $$ e = \lim_{n \to \infty} { \left(1 + {1 \over n}\right)^n }$$
Or as the value which satisfies the condition: 

For any real number $x$, $f(x) = {df\over{dx}}$

Is there an integral definition of $e$ ?
Something like: $$e = \int_0^\infty ...$$
 A: $\int_1^x \frac{1}t dt = 1$ so $e$ is the unique $x$. 
A: Would you accept $\displaystyle e = \int_0^\infty \frac1{\lfloor x\rfloor!} \,dx$?
A: I think it's worth pointing out the link between the examples you proffered.
Note $e^t$ is the solution to the differential equation $dx/dt = x, x(0) = 1$. Thus, $x(1) = e$.
Now, try to approximate the solution to this ODE using the forward Euler method with step size $1/h$ to estimate $x(1)$. The first step yields $x(1/h) \approx x(0) + (1/h)x(0) = (1 + 1/h)x(0) = 1 + 1/h$. The second step yields $x(2/h) \approx x(1/h) + (1/h)(x(1/h) = (1+1/h)x(1/h) = (1+1/h)^2$. Taking $h$ steps yields $x(1) \approx (1 + 1/h)^h$. Using consistent approximations results, you get that, as the step size $1/h$ goes to $0$, $x(1) \rightarrow e$.
Now, note that $\frac{d^2x}{dt^2} = \frac{d}{dt} \left( dx/dt \right) = \frac{d}{dt} x$. (This follows just by plugging in the equality $dx/dt = x$.) Inducting, we get that $\frac{d^nx}{dt^n} = x$ for all $n \in \mathbb{N}$, and the convergent sum result is simply a direct application of Taylor's theorem.
Basically, my point is that these equivalent definitions are not coincidence, and hint at a lot of neat results.
