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I'm sure that's a coincidence, but the Laplace transform of $1/\Gamma(x)$ at $s=1$ turns out to be pretty close to the inverse of the Golden ratio:

$$F(1)=\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx=0.61985841414477344973$$

Can we prove analytically that: $$\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx>\frac{1}{\phi}$$

The continued fraction of this number also starts very beautifully:

$$F(1)=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\dots}}}}}}}}}}=$$

$$=[0; 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 7, 2, 1, 1, 1, 2, 6, 1, 4, 7, 1, 3, 1, 1, 6, 1,\dots]$$

The question has no practical application, but everyone here loves Golden ratio questions, right?


I'm sure there is no closed form for this integral, but if there is some useful transformation, I would like to see it as well.


If we replace every partial quotient in the CF after the first 5 2's by $2$, we obtain a great approximation to the integral:

$$F(1) \approx \frac{1}{41} \left(24+\sqrt{2}\right)$$

So, the more complicated question would be:

Can we prove analytically that: $$\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx<\frac{1}{41} \left(24+\sqrt{2}\right)$$

The difference is about $4 \cdot 10^{-7}$.

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  • $\begingroup$ I think we may exploit the log-convexity of the $\Gamma$ function, but we have to be pretty careful in order to achieve such accuracy goal. $\endgroup$ – Jack D'Aurizio Jan 17 '17 at 23:10
  • $\begingroup$ How about product form of gamma function? $\endgroup$ – Simply Beautiful Art Jan 18 '17 at 12:23
  • $\begingroup$ @SimpleArt, I tried some partial products of the Weierstrass product for Gamma function, but even though they are integrated in closed form, the resulting expressions don't really tell me anything, except for the fact that the integral is less than $1$. $\endgroup$ – Yuriy S Jan 18 '17 at 12:49
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    $\begingroup$ One can use rigorous numerical integration provided by the arblib to prove/verify the lower bound. In sage you can try CBF.integral(lambda x, _: (e^(-x))/gamma(x), 1/10000000, 10000). The result is [0.619858414145 +/- 3.12e-13]. $\endgroup$ – ablmf Nov 29 '19 at 15:29
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    $\begingroup$ Split the integral over 3 segments $(0,10^{-10})$ and $(10^{-10},10^4)$ and $(10^4,\infty)$. Use rigorous numeric integral to bound each part, and we get an upper bound. [0.619858867374953 +/- 6.80e-16] $\endgroup$ – ablmf Nov 29 '19 at 16:27
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Essentially, this can be proved by dividing the integral interval into $(0,1/100)$, $(1/100,10000)$ and $(10000,\infty)$ and bound the integrand $e^{-x}/ \Gamma(x)$ with simpler functions on each part.

For the lower bound we don't really need to get into the details. There are software that can automatically compute guaranteed numeric upper and lower bounds of integrals. This type of methods are called validated numerics. One such software is arblib. You can access it through SageMath.

The following code proves the lower bound.

low=RBF(CBF.integral(lambda x, _: (e^(-x))/gamma(x), 1/100, 5))
phi=RBF((1+sqrt(5))/2)
low > 1/phi

The output is True.

For the upper bound, we use the following inequalities $$ e^{-x}/\Gamma(x) \le x, \qquad x \in (0,1/100) $$ and $$ e^{-x}/\Gamma(x) \le e^{-x}, \qquad x \in (10000,\infty) $$ So the following code proves the upper bound.

p1=RBF(CBF.integral(lambda x, _: x, 0, 1/100))
p2=RBF(CBF.integral(lambda x, _: (e^(-x))/gamma(x), 1/100,10000))
p3=RBF(integral(e^(-x), x,10000,oo))
up=RBF(1/41*(24+sqrt(2)))
up > p1+p2+p3

The output is again True.


If one want an completely analytic proof, for the lower bound one can use the inequality $$ f(x):=e^{-x}/\Gamma(x) \ge f_{1,2,5}(x) $$ where $f_{1,2,5}(x)$ is the Pade approximant of order $(1,2)$ at $x=5$. Since $f_{1,2,5}(x)$ is a rational function, one can do integrate it on $(1/100,5)$ exactly and get an analytic lower bound.

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    $\begingroup$ By nature, one can always prove an approximation of an integral by dividing it into pieces. The question is asking for an analytic proof. $\endgroup$ – JoshuaZ Dec 5 '19 at 17:45

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