Gamma-Distribution-like integral I have numerically checked this result and although it doesn't hold to a significant number of decimal places I believe this result is true:
$$\Large \int_0^\infty \frac{x^x e^{-x}}{\Gamma(x+2)}\text{d}x = 1$$
This only vaguely resembles a Gamma Distribution, so I do not see how to explain it using distributions.
I would imagine complex analysis is the way to go with such an integral but I have no idea where to even begin.
I tried using Stirling's (Convergent) Approximation but given how complicated the product expansion is in terms of exponentiated inverted rising factorials, I don't think that is a very nice method.
 A: Let $\gamma$ be Hankle's contour. It is well-known that:
$$\dfrac{1}{\Gamma(z)}=\dfrac{i}{2\pi}\int_{\gamma}(-t)^{-z}e^{-t}{\rm d}t$$
STEP 1: Considering the integral:
\begin{align*}
J(a)& =\int_{0}^{\infty}\dfrac{x^xe^{-ax}}{\Gamma(x+1)}{\rm d}x\\
& =\dfrac{i}{2\pi}\int_0^{\infty}e^{-ax}{\rm d}x\int_{\gamma}(-t)^{-x-1}e^{-tx}{\rm d}t\\
& =-\dfrac{i}{2\pi}\int_{\gamma}\dfrac{{\rm d}t}{t(a+t+\log(-t))}\\
& =-\dfrac{1}{1+W_{-1}(-e^{-a})},
\end{align*}
where $W_{-1}(z)$ is Lambert W function. The last equation is right due to residue theorem.
STEP 2: Hence \begin{align*}
I & = e\int_{0}^{\infty}{\rm d}x\int_{1}^{\infty}\dfrac{x^xe^{-a(1+x)}}{\Gamma(1+x)}{\rm d}a\\
& =e\int_1^{\infty}e^{-a}J(a){\rm d}a\\
& =-e\int_1^{\infty}\dfrac{e^{-a}}{1+W_{-1}(-e^{-a})}{\rm d}a\\
& =-e\int_0^{-1/e}\dfrac{{\rm d}t}{1+W_{-1}(t)}\\
& =e\int_{-\infty}^{-1}e^W_{-1}{\rm d}W=1
\end{align*}
The last equation is right due to $W'_{-1}(z)(1+W_{-1}(z))e^{W_{-1}(z)}=1.$
Corollary：$$\int_0^{\infty}\frac{x^{x+p-1}e^{-x}}{\Gamma(x+p+1)}dx=\frac{1}{p}.$$
Here Proposition 5.5 at Page 27
The same question has been discussed on MathOverflow, see here
