I want to evaluate the following definite integral.
$$\int_0^\infty\frac{x^n}{n!}dn$$
Where we have $n!=\Gamma(n+1)=\int_0^\infty t^ne^{-t}dt$ so that we can have $n\in\mathbb{R}$.
I don't think there is a solution to this, but I do note that
$$\int_0^\infty\frac{x^n}{n!}dn\sim\sum_{n=0}^\infty\frac{x^n}{n!}=e^x$$
Also, we define $0^0=1$.
The context to this problem is that I want to see if
$$\int_0^\infty\frac{x^nf^{(n)}(a)}{n!}dn=\sum_{n=0}^\infty\frac{x^nf^{(n)}(a)}{n!}$$
A modification to Taylor's theorem for fractional calculus.
Under my conditions, it is obvious that the two are equal for $x=0$, but even for the seemingly trivial case of $x=1$, I don't know how to solve the integral.
WolframAlpha provides a series representation for the indefinite integral ($x=1$), but I am unsure if I can even use it.
Feel free to tackle just the $x=1$ case. From graphing, it seems very possible that $\int_0^\infty\frac1{n!}dn=e$, but I'm not fully sure.
From wolframalpha, I have found that $\int_0^\infty\frac1{n!}d\approx2.2665$.
EDIT
Numerical calculations say $\int_{-\gamma}^\infty\frac1{n!}dn\approx2.70907$ where $\gamma$ is the Euler-Mascheroni constant. It seems very close to $e$...
I doubt there's a closed form, but you can write it as
$$ \int_0^1 \sum_{j=0}^\infty \dfrac{x^{a+j}}{\Gamma(a+j+1)}\; da = \int_0^1 e^x \left(1 - \dfrac{a \Gamma(a,x)}{\Gamma(a+1)}\right)\; da $$
Numerically, the value at $x=1$ is approximately $2.266534508$.