Integral involving Gamma function: density of Kendall-Ressel family of distributions

Question

I came across the following function when reading the famous paper of Letac and Mora "Natural exponential families with cubic variance functions", i.e., $$f(x) = \frac{x^x e^{-x}}{\Gamma(x+2)}$$ for $x \ge 0$. Proposition 5.5 there showed via a transform of a Levy process that $f$ is a density on $(0,\infty)$ without the need to compute the integral $$\int_{0}^{\infty} f(x)dx$$ In fact, $f$ is called Kendall-Ressel density.

Tonight, I used Mathematica to compute the above integral but Mathematica did not give the answer $1$. Instead, it was stuck and returned the orginal integral. So, I guess a direct computation is not trivial!?

I am curious on how to directly show that $$\int_{0}^{\infty} f(x)dx = 1$$ Any suggestions? Thank you.

Update on June 21, 2018: to follow up on Nemo's solution: is there a general formula or recursive formula for $$I(b,\alpha) = \int_{C} \frac{t^{-1}dt}{(t + \ln(-t)+b)^\alpha},$$ where $\alpha$ is a positive constant? Say, $\alpha$ is a natural number.

Answer 1

According to Hankel's formula $$ \frac{1}{\Gamma(z)}=\frac{i}{2\pi}\int_C(-t)^{-z}e^{-t}dt, $$ where $C$ is Hankel contour. So $$ \frac{x^x}{\Gamma(x+1)}=\frac{i}{2\pi}\int_C(-t)^{-x-1}e^{-xt}dt,\quad x>0. $$ Consider the integral $$ I(b)=\int_0^\infty \frac{x^xe^{-bx}}{\Gamma(x+1)}dx=\int_0^\infty e^{-bx}dx\frac{i}{2\pi}\int_C(-t)^{-x-1}e^{-xt}dt. $$ Changing the order of integration and calculating the integral over $x$ we get $$ I(b)=-\frac{i}{2\pi}\int_C\frac{dt}{t(t+\ln(-t)+b)}. $$ This integral can be easily calculated using residue theory $$ I(b)=-\frac{1}{1+W_{-1}(-e^{-b})}, $$ where $W_{-1}(z)$ is the Lambert W-function, satisfying $W_{-1}(z)e^{W_{-1}(z)}=z$, which has the derivative $$ W_{-1}'(z)=\frac{e^{-W_{-1}(z)}}{1+W_{-1}(z)}.\tag{*} $$

Now we write the initial integral as follows \begin{align} \int_0^\infty \frac{x^x e^{-x}}{\Gamma(x+2)}dx&=e\int_1^\infty e^{-b}I(b)db\\ &=-e\int_1^\infty e^{-b}\frac{1}{1+W_{-1}(-e^{-b})}db\\ &=-e\int_0^{1/e} \frac{ds}{1+W_{-1}(-s)}\\ &=e\int_0^{-1/e} \frac{ds}{1+W_{-1}(s)}\\ &=e\int_{-\infty}^{-1} e^{W}d{W}=e\cdot 1/e=1. \end{align} Here we used the equation $(*)$ to write $\frac{1}{1+W_{-1}(s)}=e^{W_{-1}(s)}W_{-1}'(s)$ and also the facts $W_{-1}(-1/e)=-1~$, $W_{-1}(0)=-\infty$.