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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.

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  • $\begingroup$ The original post is here in case you wonder how this appeared. $\endgroup$
    – Zacky
    Commented Dec 12, 2019 at 9:52

2 Answers 2

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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

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  • $\begingroup$ $\int_0^{\infty } \frac{x^x \exp (-a x)}{\Gamma (x+1)} \, dx=-\frac{1}{1+W(-\exp (-a))}$ is not true,I checked numerically ? $\endgroup$ Commented Jan 19, 2020 at 10:59
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    $\begingroup$ @Mariusz Iwaniuk, I believe one has been a little sloppy here. When dealing with the Lambert W function one should always specify which branch of the function one is dealing with. In this case it is the $k = -1$ branch. Thus $$\int_0^\infty \frac{x^x e^{-ax}}{\Gamma (x + 1)} \, dx = -\frac{1}{1 + \operatorname{W}_{-1} (-e^{-a})}$$ for $a > 1$. $\endgroup$
    – omegadot
    Commented Jan 20, 2020 at 1:22
  • $\begingroup$ Sorry @Mariusz Iwaniuk, I meant to say the solver has been a little sloppy in not specifying which branch of the Lambert W function they were considering, not you. $\endgroup$
    – omegadot
    Commented Jan 20, 2020 at 7:12
  • $\begingroup$ @Mariusz Iwaniuk Sorry for my fault. Thanks $\endgroup$ Commented Jan 20, 2020 at 9:40
  • $\begingroup$ Thanks for correcting an answer. $\endgroup$ Commented Jan 20, 2020 at 16:02
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I cannot access the paper in the linked closed question.

Considering your problem, let us write $$I=\int_{0}^\infty \frac{x^x e^{-x}}{\Gamma(x+2)}\,dx=\int_0^{10^k} \frac{x^x e^{-x}}{\Gamma(x+2)}\,dx+\int_{10^k}^\infty \frac{x^x e^{-x}}{\Gamma(x+2)}\,dx=I_{1k}+I_{2k}$$ $I_{1k}$ would be computed numerically.

Concerning the second one, for large value of $10^k=p$, using Stirling approximation and continuing with Taylor series, we have $$\frac{x^x e^{-x}}{\Gamma(x+2)}=\frac{1}{\sqrt{2 \pi x^3}} \left(1-\frac{13}{12 x}+\frac{313}{288 x^2}-\frac{56201}{51840 x^3}+O\left(\frac{1}{x^4}\right)\right)$$ from which $$\int_p^\infty\frac{x^x e^{-x}}{\Gamma(x+2)}\,dx=\sqrt{\frac{2}{\pi\,p }} \left(1-\frac{13}{36 p}+\frac{313}{1440 p^2}-\frac{56201}{362880 p^3}+O\left(\frac{1}{p^4}\right) \right)$$ Now, computing for a few values of $k$ $$\left( \begin{array}{cccc} k & I_{1k} & I_{2k} & I_{1k}+ I_{2k} \\ 1 & 0.75628589652513868756 & 0.24371129421828760677 & 0.99999719074342629434 \\ 2 & 0.92049794687117372168 & 0.07950205303351501669 & 0.99999999990468873837 \\ 3 & 0.97477778061127925239 & 0.02522221938871771364 & 0.99999999999999696603 \\ 4 & 0.99202144249960996668 & 0.00797855750039003538 & 1.00000000000000000206 \\ 5 & 0.99747687658923688302 & 0.00252312341076311767 & 1.00000000000000000069 \\ 6 & 0.99920211572732194172 & 0.00079788427267805849 & 1.00000000000000000022 \end{array} \right)$$

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